{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $MNP$ c\u00f3 $MN < MP$. $I$ l\u00e0 m\u1ed9t \u0111i\u1ec3m n\u1eb1m tr\u00ean $MP$ sao cho $MI < MN$. $K$ l\u00e0 \u0111i\u1ec3m n\u1eb1m gi\u1eefa $N$ v\u00e0 $I$. Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $KN + KM < MN$","B. $MP - MN > NP$","C. $KN + KI > MN - MI$","D. $\\widehat{MNI} > \\widehat{MIN}$"],"hint":"V\u1eadn d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c, h\u1ec7 qu\u1ea3 v\u00e0 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00e0 h\u1ec7 qu\u1ea3 cho c\u00e1c tam gi\u00e1c $MKN; MNP; MIN$ \u0111\u1ec3 x\u00e9t c\u00e1c \u0111\u00e1p \u00e1n A, B, C <br\/> <b> B\u01b0\u1edbc 2: <\/b> S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{MIN}$ \u0111\u1ec3 so s\u00e1nh $\\widehat{MNI}$ v\u00e0 $\\widehat{MIN}$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D101.png' \/><\/center> <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{MKN}$, ta c\u00f3: <br\/> $KM + KN > MN$ $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A sai <\/b> <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{MNP}$, ta c\u00f3: <br\/> $MP - MN < NP$ $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B sai <\/b> <br\/> $\\blacktriangleright$ V\u00ec \u0111i\u1ec3m $K$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $N$ v\u00e0 $I$ n\u00ean ta c\u00f3 $KN + KI = NI$ <br\/> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{MIN}$, ta c\u00f3: <br\/> $NI > MN - MI$ hay $KN + KI > MN - MI$ $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{MNI}$ c\u00f3 $MI < MN$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow$ $\\widehat{MNI} < \\widehat{MIN}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n D sai <\/b> <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> ","column":2}],"id_ques":2041},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 s\u1ed1 \u0111o c\u00e1c c\u1ea1nh l\u00e0 s\u1ed1 nguy\u00ean. Bi\u1ebft $AB = 6cm, BC = 1cm$. Chu vi tam gi\u00e1c $ABC$ l\u00e0: ","select":["A. $11cm$","B. $12cm$","C. $13cm$","D. $14cm$"],"hint":"V\u1eadn d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00e0 h\u1ec7 qu\u1ea3","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ecm s\u1ed1 \u0111o c\u1ea1nh $AC$ b\u1eb1ng c\u00e1ch \u00e1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00e0 h\u1ec7 qu\u1ea3 <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh chu vi tam gi\u00e1c $ABC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00e0 h\u1ec7 qu\u1ea3 cho $\\triangle{ABC}$, ta c\u00f3: <br\/> $AB - BC < AC < AB + BC$ <br\/> $6 - 1 < AC < 6 + 1$ $\\Rightarrow$ $5 < AC < 7$ <br\/> V\u00ec s\u1ed1 \u0111o c\u00e1c c\u1ea1nh l\u00e0 s\u1ed1 nguy\u00ean n\u00ean $AC = 6cm$ <br\/> Chu vi tam gi\u00e1c $ABC$ l\u00e0: $6 + 1 + 6 = 13(cm)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: $C. 13cm$ <\/span> ","column":2}],"id_ques":2042},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $IHL$ vu\u00f4ng t\u1ea1i $I$, $M$ v\u00e0 $N$ l\u00e0 hai \u0111i\u1ec3m l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean $IH$ v\u00e0 $IL$. So s\u00e1nh n\u00e0o sau \u0111\u00e2y l\u00e0 <b> sai<\/b>? ","select":["A. $HN < HL$","B. $HN > IH$","C. $MN < ML$","D. $MN < IN$"],"hint":"S\u1eed d\u1ee5ng m\u1ed1i quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c \u0111\u1ec3 x\u00e9t t\u1eebng \u0111\u00e1p \u00e1n","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D102.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{HIN}$ vu\u00f4ng t\u1ea1i $I$ c\u00f3 $\\widehat{N_{1}}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $N$, n\u00ean: <br\/> $\\widehat{N_{1}} > \\widehat{I} = 90^{o}$ <br\/> $\\triangle{NHL}$ c\u00f3 $\\widehat{N_{1}} > 90^{o}$ <br\/> $\\Rightarrow$ $HL > HN$ (trong tam gi\u00e1c t\u00f9, c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c t\u00f9 l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ $\\triangle{IHN}$ vu\u00f4ng t\u1ea1i $I$ n\u00ean c\u1ea1nh huy\u1ec1n $HN$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t <br\/> Hay $HN > IH$ $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ $\\triangle{IMN}$ vu\u00f4ng t\u1ea1i $I$ c\u00f3 $\\widehat{MNL}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $N$, n\u00ean: <br\/> $\\widehat{MNL} > \\widehat{I} = 90^{o}$ <br\/> $\\triangle{MNL}$ c\u00f3 $\\widehat{MNL} > 90^{o}$ n\u00ean $ML > MN$ (trong tam gi\u00e1c t\u00f9, c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c t\u00f9 l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t) <br\/> $\\Rightarrow$ \u0110\u00e1p \u00e1n C \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ $\\triangle{IMN}$ vu\u00f4ng t\u1ea1i $I$ n\u00ean c\u1ea1nh huy\u1ec1n $MN$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t <br\/> $\\Rightarrow$ $MN > IN$ $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n D sai <\/b> <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n sai l\u00e0: D <\/span> ","column":2}],"id_ques":2043},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 trung tuy\u1ebfn $AM$. G\u1ecdi $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c. K\u1ebft lu\u1eadn n\u00e0o sau \u0111\u00e2y l\u00e0 <b> sai <\/b>? ","select":["A. $AG = 2GM$","B. $GM = \\dfrac{1}{2}AM$","C. $GM = \\dfrac{1}{3}AM $","D. $AG = \\dfrac{2}{3}AM $"],"hint":"S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D103.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{ABC}$ n\u00ean $AG = \\dfrac{2}{3}AM$ (t\u00ednh ch\u1ea5t) (1) <b> (\u0110\u00e1p \u00e1n D \u0111\u00fang) <\/b> <br\/> $\\blacktriangleright$ V\u00ec $AG = \\dfrac{2}{3}AM \\Rightarrow GM = \\dfrac{1}{3}AM$ (2) <b> (\u0110\u00e1p \u00e1n C \u0111\u00fang, \u0111\u00e1p \u00e1n B sai) <\/b> <br\/> $\\blacktriangleright$ T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\dfrac{AG}{GM} = \\dfrac{2}{3}AM : \\dfrac{1}{3}AM = 2$ <br\/> Hay $AG = 2GM$ <b> (\u0110\u00e1p \u00e1n A \u0111\u00fang) <\/b> <br\/> <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n sai l\u00e0: B <\/span> ","column":2}],"id_ques":2044},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $MNP$ vu\u00f4ng t\u1ea1i $M$ c\u00f3 tr\u1ef1c t\u00e2m $H$ th\u00ec khi \u0111\u00f3: ","select":["A. $H$ l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb $M$ xu\u1ed1ng $NP$","B. $H$ n\u1eb1m trong tam gi\u00e1c ","C. $H$ tr\u00f9ng v\u1edbi $M$ ","D. $H$ n\u1eb1m ngo\u00e0i tam gi\u00e1c"],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D104.png' \/><\/center> <br\/> $\\triangle{MNP}$ vu\u00f4ng t\u1ea1i $M$ c\u00f3 $MN \\perp MP$ t\u1ea1i $M$ <br\/> Hay $NM; PM$ ch\u00ednh l\u00e0 hai \u0111\u01b0\u1eddng xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh $N$ v\u00e0 \u0111\u1ec9nh $P$ ch\u00fang c\u1eaft nhau t\u1ea1i $M$ <br\/> M\u00e0 ba \u0111\u01b0\u1eddng cao c\u1ee7a m\u1ed9t tam gi\u00e1c c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m n\u00ean tr\u1ef1c t\u00e2m $H$ tr\u00f9ng v\u1edbi \u0111i\u1ec3m $M$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> ","column":2}],"id_ques":2045},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$. K\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AM$ c\u1ee7a $\\widehat{BAC}$ ($M \\in BC$). $I$ l\u00e0 m\u1ed9t \u0111i\u1ec3m b\u1ea5t k\u00ec tr\u00ean $AM$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y sai? ","select":["A. $IB$ v\u00e0 $IC$ l\u00e0 hai \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{ABC}$","B. $AM$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BC$ ","C. $\\triangle{ABM} = \\triangle{ACM}$ ","D. $\\triangle{BIM} = \\triangle{CIM}$ "],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D105.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3 $AM$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $\\widehat{A}$ <br\/> $I$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec thu\u1ed9c $AM$ n\u00ean $IB, IC$ kh\u00f4ng ph\u1ea3i l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{ABC}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A sai <\/b> <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$, $AM$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $\\widehat{A}$ <br\/> $\\Rightarrow$ $AM$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BC$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABM}$ v\u00e0 $\\triangle{ACM}$ c\u00f3: <br\/> $\\begin{cases} AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\\\ \\widehat{BAM} = \\widehat{CAM} (gt) \\\\ AM \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABM} = \\triangle{ACM}$ (c.g.c) $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta \u0111\u01b0\u1ee3c: $\\triangle{BIM} = \\triangle{CIM}$ (c.g.c) $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n D \u0111\u00fang <\/b> <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n sai l\u00e0: A <\/span> ","column":1}],"id_ques":2046},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho g\u00f3c nh\u1ecdn $xOy$. Tr\u00ean c\u1ea1nh $Ox$ l\u1ea5y hai \u0111i\u1ec3m $A; B$ sao cho $A$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $O$ v\u00e0 $B$. Tr\u00ean c\u1ea1nh $Oy$ l\u1ea5y hai \u0111i\u1ec3m $C; D$ sao cho $C$ n\u1eb1m gi\u1eefa $O$ v\u00e0 $D$. Khi \u0111\u00f3 $AB + CD < AD + BC$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG","B. SAI "],"hint":"\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c","explain":" <span class='basic_left'> <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AD$ v\u00e0 $BC$ <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho tam gi\u00e1c $ABI$ v\u00e0 $CDI$ <br\/> <b> B\u01b0\u1edbc 2:<\/b> C\u1ed9ng theo v\u1ebf c\u1ee7a hai b\u1ea5t \u0111\u1eb3ng th\u1ee9c tr\u00ean \u0111\u1ec3 c\u00f3 b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u1ea7n t\u00ecm. <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D108.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AD$ v\u00e0 $BC$. <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{ABI}$, ta \u0111\u01b0\u1ee3c: <br\/> $AB < AI + BI$ (1) <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{CDI}$, ta \u0111\u01b0\u1ee3c: <br\/> $CD < CI + DI$ (2) <br\/> $\\blacktriangleright$ C\u1ed9ng theo t\u1eebng v\u1ebf c\u1ee7a (1) v\u1edbi (2) ta \u0111\u01b0\u1ee3c: <br\/> $AB + CD < AI + BI + CI + DI = AD + BC$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":1}],"id_ques":2047},{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","4"]],"list":[{"point":10,"img":"","ques":"Cho tam gi\u00e1c $MNP$ c\u00f3 $MN < MP$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $NP$ l\u1ea5y \u0111i\u1ec3m $R$ sao cho $NR = NM$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $PN$ l\u1ea5y \u0111i\u1ec3m $Q$ sao cho $PQ = PM$. N\u1ed1i $M$ v\u1edbi $R$, $M$ v\u1edbi $Q$. H\u00e3y ch\u1ecdn <b> nh\u1eefng <\/b> kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang.","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u00e0 t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c","column":2,"number_true":2,"select":["A. $\\widehat{RMN} > \\widehat{PMQ}$ ","B. $\\widehat{RMN} < \\widehat{PMQ}$ ","C. $MR = MQ$ ","D. $MQ > MR$"],"explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh $\\widehat{MPN}$ v\u00e0 $\\widehat{MNP}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i so s\u00e1nh $\\widehat{RMN}$ v\u1edbi $\\widehat{MNP}$; $\\widehat{PMQ}$ v\u1edbi $\\widehat{MPN}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u1eeb b\u01b0\u1edbc 1 v\u00e0 b\u01b0\u1edbc 2 so s\u00e1nh $\\widehat{RMN}$ v\u00e0 $\\widehat{PMQ}$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> So s\u00e1nh $MQ$ v\u00e0 $MR$ d\u1ef1a v\u00e0o quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{MRQ}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D109.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{MNP}$ c\u00f3 $MN < MP$ n\u00ean $\\widehat{MPN} < \\widehat{MNP}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{MNP}$) (1) <br\/> $\\blacktriangleright$ $\\triangle{MNR}$ c\u00f3 $NP = NM$ n\u00ean $\\triangle{MNR}$ c\u00e2n t\u1ea1i $N$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ $\\widehat{MRN} = \\widehat{RMN}$ (t\u00ednh ch\u1ea5t) <br\/> M\u1eb7t kh\u00e1c: $ \\widehat{MRN} + \\widehat{RMN} = \\widehat{MNP}$ (t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i c\u1ee7a $\\triangle{MRN}$) <br\/> N\u00ean $\\widehat{MRN} = \\widehat{RMN} = \\dfrac{1}{2}\\widehat{MNP}$ (2) <br\/> T\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3: $\\widehat{PMQ} = \\widehat{PQM} = \\dfrac{1}{2}\\widehat{MPN}$ (3) <br\/> $\\blacktriangleright$ T\u1eeb (1), (2), (3) $\\Rightarrow$ $\\widehat{RMN} > \\widehat{PMQ}$ <b> (\u0110\u00e1p \u00e1n A \u0111\u00fang, B sai) <\/b> <br\/> $\\blacktriangleright$ $\\triangle{MRQ}$ c\u00f3 $\\widehat{MRQ} > \\widehat{MQP}$ (v\u00ec $\\widehat{RMN} > \\widehat{PMQ}$) <br\/> $\\Rightarrow$ $MQ > MR$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{MRQ}$) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C sai, D \u0111\u00fang <\/b> <br\/> <span class='basic_pink'> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A, D <\/span> "}],"id_ques":2048},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $B$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y m\u1ed9t \u0111i\u1ec3m $H$ sao cho $\\widehat{ACH} = \\dfrac{1}{3}\\widehat{ACB}$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BC$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $BK = BH$. T\u00ednh $\\widehat{AKH}$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{AKH} = $ _input_$^{o}$ ","hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $KH \\perp AC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $CH \\perp AK$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u1eeb b\u01b0\u1edbc 1 v\u00e0 b\u01b0\u1edbc 2 suy ra \u0111\u01b0\u1ee3c hai g\u00f3c c\u00f3 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng vu\u00f4ng g\u00f3c $\\widehat{AKH}$ v\u00e0 $\\widehat{ACH}$ b\u1eb1ng nhau <br\/> T\u1eeb \u0111\u00f3 t\u00ednh s\u1ed1 \u0111o $\\widehat{ACH}$ r\u1ed3i $\\Rightarrow$ s\u1ed1 \u0111o $\\widehat{AKH}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D106.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $M$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $KH$ v\u00e0 $AC$, $N$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $CH$ v\u1edbi $AK$ <br\/> $\\triangle{ABC}$ vu\u00f4ng c\u00e2n t\u1ea1i $B$ n\u00ean $\\widehat{ACB} = 45^{o}$ <br\/> $\\triangle{HKB}$ vu\u00f4ng t\u1ea1i $B$ c\u00f3 $BH = BK$ (gt) n\u00ean l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n <br\/> $\\Rightarrow$ $\\widehat{HKB} = 45^{o}$ <br\/> X\u00e9t $\\triangle{KMC}$ c\u00f3 $\\widehat{KMC} = 180^{o} - (\\widehat{MKC} + \\widehat{MCK}) = 180^{o} - (45^{o} + 45^{o}) = 90^{o}$ <br\/> V\u1eady $KM \\perp AC$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AKC}$ c\u00f3 $AB$ v\u00e0 $KM$ l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1eaft nhau t\u1ea1i $H$ <br\/> $\\Rightarrow$ $CH$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao trong 1 tam gi\u00e1c) <br\/> Hay $CH \\perp AK$ <br\/> $\\blacktriangleright$ Ta c\u00f3 $\\widehat{AKH} = \\widehat{ACH}$ (g\u00f3c c\u00f3 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng vu\u00f4ng g\u00f3c) <br\/> M\u00e0 $\\widehat{ACH} = \\dfrac{1}{3}\\widehat{ACB} = \\dfrac{1}{3}.45^{o} = 15^{o}$ <br\/> N\u00ean $\\widehat{AKH} = 15^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $15$ <\/span> <br\/> <span class='basic_green'><i> Nh\u1eadn x\u00e9t: T\u1eeb t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c c\u00f9ng \u0111i qua 1 \u0111i\u1ec3m (tr\u1ef1c t\u00e2m) ta suy ra: Trong m\u1ed9t tam gi\u00e1c, \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua m\u1ed9t \u0111\u1ec9nh v\u00e0 tr\u1ef1c t\u00e2m c\u0169ng l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng cao. \u0110\u00e2y l\u00e0 m\u1ed9t c\u00e1ch m\u1edbi \u0111\u1ec3 ch\u1ee9ng minh hai \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c. <\/i> <\/span> "}],"id_ques":2049},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, $\\widehat{A} = 30^{o}, BC = 2$. Tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $AD = \\sqrt{2}$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $ABD$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{ABD} = $ _input_$^{o}$ ","hint":"Trong $\\triangle{ABC}$ v\u1ebd $\\triangle{EBC}$ vu\u00f4ng c\u00e2n t\u1ea1i $E$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Trong $\\triangle{ABC}$ v\u1ebd $\\triangle{EBC}$ vu\u00f4ng c\u00e2n t\u1ea1i $E$ sau \u0111\u00f3 t\u00ednh $EB$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{BAE}$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\widehat{BAE} = \\widehat{CAE}$ th\u00f4ng qua ch\u1ee9ng minh hai tam gi\u00e1c b\u1eb1ng nhau <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{ABE}$ d\u1ef1a v\u00e0o g\u00f3c $ABC$ v\u00e0 g\u00f3c $EBC$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh $\\widehat{ABD}$ th\u00f4ng qua ch\u1ee9ng minh $\\triangle{ABD} = \\triangle{BAE}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/H7kiemtrachuong3/lv3/img\/H7kiemtrachuong3_D107.png' \/><\/center> <br\/> $\\blacktriangleright$ Trong $\\triangle{ABC}$ v\u1ebd tam gi\u00e1c $EBC$ vu\u00f4ng c\u00e2n t\u1ea1i $E$ <br\/> $\\triangle{EBC}$ vu\u00f4ng c\u00e2n t\u1ea1i $E$ n\u00ean $\\widehat{EBC} = 45^{o}$, $EB = EC$ (t\u00ednh ch\u1ea5t) <br\/> Ta c\u00f3: $EB^2 + EC^2 = BC^2$ (\u0111\u1ecbnh l\u00fd Pitago) <br\/> $\\Rightarrow$ $2 EB^2 = 2^2$ hay $2EB^2 = 4$ <br\/> $\\Rightarrow$ $EB^2 = 2$ $\\Rightarrow$ $EB = \\sqrt{2}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BAE}$ v\u00e0 $\\triangle{CAE}$ c\u00f3: <br\/> $\\begin{cases} EB = EC (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{v\u1ebd}) \\\\ AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\\\ AE \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AEB} = \\triangle{AEC}$ (c.c.c) <br\/> $\\Rightarrow$ $\\widehat{BAE} = \\widehat{CAE} = \\dfrac{1}{2}\\widehat{BAC} = 15^{o}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\blacktriangleright$ $\\widehat{ABC} = \\widehat{ACB} = (180^{o} - \\widehat{A}) : 2$ ($\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$) <br\/> $\\Rightarrow$ $\\widehat{ABC} = \\widehat{ACB} = (180^{o} - 30^{o}) : 2 = 75^{o}$ <br\/> V\u1eady $\\widehat{ABE} = \\widehat{ABC} - \\widehat{EBC} = 75^{o} - 45^{o} = 30^{o}$ <br\/> $\\Rightarrow$ $\\widehat{ABE} = \\widehat{BAD} = 30^{o}$ <br\/> $\\blacktriangleright$ $\\triangle{ABD} = \\triangle{BAE}$ (c.g.c) v\u00ec: <br\/> $\\begin{cases} AB \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{ABE} = \\widehat{BAD} (cmt) \\\\ AD = EB = \\sqrt{2} \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{ABD} = \\widehat{BAE} = 15^{o}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $15$ <\/span> <br\/> <span class='basic_green'><i> Nh\u1eadn x\u00e9t: Trong b\u00e0i tr\u00ean ta v\u1ebd h\u00ecnh ph\u1ee5 b\u1eb1ng c\u00e1ch v\u1ebd tam gi\u00e1c vu\u00f4ng c\u00e2n $EBC$ v\u00ec trong \u0111\u1ec1 b\u00e0i cho $BC = 2; AD = \\sqrt{2}$, theo \u0111\u1ecbnh l\u00fd Pitago \u0111\u1ed9 d\u00e0i $\\sqrt{2}$ v\u00e0 $2$ ch\u00ednh l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ee7a c\u1ea1nh g\u00f3c vu\u00f4ng v\u00e0 c\u1ea1nh huy\u1ec1n c\u1ee7a m\u1ed9t tam gi\u00e1c vu\u00f4ng c\u00e2n. V\u1edbi c\u00e1ch v\u1ebd nh\u01b0 v\u1eady d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c c\u00e1c tam gi\u00e1c b\u1eb1ng nhau, t\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u01b0\u1ee3c s\u1ed1 \u0111o g\u00f3c $ABD$ <\/i> <\/span> "}],"id_ques":2050}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}