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{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["78"],["102"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho h\u00ecnh thoi $MNPQ$ trong h\u00ecnh sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_1.jpg' \/><\/center> Bi\u1ebft $\\widehat{N}-\\widehat{M}=24^o$ <br\/> T\u00ednh s\u1ed1 \u0111o c\u00e1c g\u00f3c $M$ v\u00e0 $N$ c\u1ee7a h\u00ecnh thoi. <br\/><br\/> \u0110\u00e1p \u00e1n:<\/b> <br\/> $\\widehat{M}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <br\/> $\\widehat{N}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_1.jpg' \/><\/center> Theo t\u00ednh ch\u1ea5t c\u1ee7a h\u00ecnh thoi: $\\widehat{M}+\\widehat{N}=180^o$ <br\/> M\u00e0 theo b\u00e0i cho: $\\widehat{N}-\\widehat{M}=24^o$ <br\/> Do \u0111\u00f3 ta c\u00f3 h\u1ec7: <br\/> $\\left\\{ \\begin{aligned} & \\widehat{M}+\\widehat{N}={{180}^{o}} \\\\ & \\widehat{N}-\\widehat{M}={{24}^{o}} \\\\ \\end{aligned} \\right. \\Rightarrow \\left\\{ \\begin{aligned} & \\widehat{M}=\\dfrac{180^o-24^o}{2}={{78}^{o}} \\\\ & \\widehat{N}=\\dfrac{180^o+24^o}{2}= {{102}^{o}} \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & \\widehat{M}={{78}^{o}} \\\\ & \\widehat{N}={{102}^{o}} \\\\ \\end{aligned} \\right.$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $78$ v\u00e0 $102$ <\/span> <\/span> "}]}],"id_ques":1481},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["120"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho h\u00ecnh thoi $MNPQ$ trong h\u00ecnh sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_2.jpg' \/><\/center> Bi\u1ebft $MP\\bot NQ$ t\u1ea1i $O$ v\u00e0 $NP=13\\, cm; NO=12\\, cm$ <br\/> T\u00ednh di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh thoi. <br\/><br\/> \u0110\u00e1p \u00e1n:<\/b> <br\/> $S_{MNPQ}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\, (cm^2)$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_2.jpg' \/><\/center> X\u00e9t tam gi\u00e1c vu\u00f4ng $NPO$: <br\/> $NP^2=NO^2+OP^2$ (\u0111\u1ecbnh l\u00fd Pi - ta - go) <br\/> $\\Rightarrow OP^2=NP^2-NO^2=13^2-12^2$ <br\/> $OP^2=25$ <br\/> $\\Rightarrow OP=5\\, (cm)$ <br\/> M\u00e0 $MP=2OP=10\\, (cm)$ v\u00e0 $NQ=2NO=24\\, (cm)$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi) <br\/> Di\u1ec7n t\u00edch h\u00ecnh thoi $MNPQ$ l\u00e0: <br\/> $\\dfrac{1}{2}MP.NQ$$= \\dfrac{1}{2}.24.10=120\\, (cm^2)$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $120$ <\/span> <\/span> "}]}],"id_ques":1482},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["10"],["3"],["10"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho h\u00ecnh thoi $ABCD$ v\u00e0 $O$ l\u00e0 giao c\u1ee7a hai \u0111\u01b0\u1eddng ch\u00e9o, c\u00f3 $S_{ABCD}=50\\sqrt{3}\\, cm^2$ v\u00e0 $AC=10\\, cm.$ <br\/> C\u00e2u 1: <\/b> T\u00ednh \u0111\u1ed9 d\u00e0i $BD$ v\u00e0 $AB$. <br\/><br\/> \u0110\u00e1p \u00e1n: <\/b> <br\/> $BD= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}\\, (cm)$ <br\/> $AB= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\, (cm)$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_3.jpg' \/><\/center> Theo b\u00e0i ta c\u00f3: <br\/> $S_{ABCD}=50\\sqrt{3}\\, cm^2$ <br\/> $\\Rightarrow\\dfrac{1}{2}AC.BD=50\\sqrt{3}$ <br\/> $\\Rightarrow \\dfrac{1}{2}.10.BD=50\\sqrt{3}$ <br\/> $\\Rightarrow BD=10\\sqrt{3}\\, (cm)$ <br\/> $\\Rightarrow OB= 5\\sqrt{3}\\, (cm)$ <br\/> X\u00e9t trong tam gi\u00e1c vu\u00f4ng $AOB$ c\u00f3: <br\/> $AB^2=AO^2+OB^2$ (\u0111\u1ecbnh l\u00ed Py - ta - go) <br\/> $\\Rightarrow AB^2=5^2+(5\\sqrt{3})^2$ <br\/> $\\Rightarrow AB^2=100 \\Rightarrow AB=10\\, (cm)$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 $BD=10\\sqrt{3}\\, (cm)$; $AB=10\\, (cm)$ <\/span> <\/span> "}]}],"id_ques":1483},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["120"],["60"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho h\u00ecnh thoi $ABCD$ v\u00e0 $O$ l\u00e0 giao c\u1ee7a hai \u0111\u01b0\u1eddng ch\u00e9o c\u00f3 $S_{ABCD}=50\\sqrt{3}\\, cm^2$ v\u00e0 $AC=10\\, cm$ <br\/> C\u00e2u 2: <\/b> T\u00ednh s\u1ed1 \u0111o c\u00e1c g\u00f3c c\u1ee7a h\u00ecnh thoi. <br\/><br\/> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{A}=\\widehat{C}= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <br\/> $\\widehat{B}=\\widehat{D}= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_3.jpg' \/><\/center> Theo c\u00e2u 1, ta \u0111\u00e3 t\u00ednh \u0111\u01b0\u1ee3c: $AB=10\\, cm$ v\u00e0 $OA=5\\, cm$ <br\/> Ta c\u00f3 tam gi\u00e1c vu\u00f4ng $AOB$ = tam gi\u00e1c vu\u00f4ng $COB$. T\u00ednh ch\u1ea5t C\u1ea1ch, G\u00f3c, C\u1ea1nh. <br\/> Do \u0111\u00f3 c\u1ea1nh BC = AB. N\u00ean tam gi\u00e1c $ABC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u v\u00ec c\u00f3 3 c\u1ea1nh b\u1eb1ng nhau v\u00e0 \u0111\u1ec1u b\u1eb1ng 10(cm) <br\/> $\\Rightarrow \\widehat{BAO}=60^o$ <br\/> $\\Rightarrow \\widehat{A} = 2\\widehat{BAO}=120^o$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi) <br\/> M\u00e0 $\\widehat{A}+\\widehat{B}=180^o$ <br\/> Do \u0111\u00f3 suy ra $\\widehat{B}=60^o$ <br\/> V\u1eady $\\widehat{A}=\\widehat{C}=120^o$; $\\widehat{B}=\\widehat{D}=60^o$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 $\\widehat{A}=\\widehat{C}=120^o$; $\\widehat{B}=\\widehat{D}=60^o$ <\/span> <\/span> "}]}],"id_ques":1484},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["60"],["60"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho tam gi\u00e1c $ABC$ \u0111\u1ec1u. G\u1ecdi $M\\in BC$, $E$ v\u00e0 $F$ l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c k\u1ebb t\u1eeb $M$ \u0111\u1ebfn $AB, AC.$ G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AM, D$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$ <br\/> C\u00e2u 1: <\/b> T\u00ednh s\u1ed1 \u0111o c\u00e1c g\u00f3c $DIE$ v\u00e0 g\u00f3c $DIF.$ <br\/><br\/> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{DIE}= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <br\/> $\\widehat{DIF}= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_4.jpg' \/><\/center> Ta c\u00f3: $\\Delta AEM$ vu\u00f4ng t\u1ea1i $E$ (gi\u1ea3 thi\u1ebft) <br\/> $EI$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn n\u00ean $IE=IA=IM$ <br\/> $\\Rightarrow \\widehat{EIM}=\\widehat{EAI}+\\widehat{AEI}=2\\widehat{EAI}$ (1) <br\/> Ta c\u00f3 $\\Delta ABC$ \u0111\u1ec1u, $DB = DC$ <br\/> Suy ra $AD\\bot BD$ v\u00e0 $\\widehat{BAD} = \\widehat{CAD} = 30^o$ <br\/> Tam gi\u00e1c $ADM$ vu\u00f4ng t\u1ea1i $D$ c\u00f3 $DI$ l\u00e0 trung tuy\u1ebfn n\u00ean $ID=IA=IM; \\widehat{I_1}=2\\widehat{A_1}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{EIM}-\\widehat{I_1}=2(\\widehat{EAI}-\\widehat{A_1})$ <br\/> $\\Rightarrow \\widehat{I_2}=2\\widehat{A_2}=60^o$ <br\/> V\u1eady $\\widehat{DIE}=60^o$ <br\/> T\u01b0\u01a1ng t\u1ef1 nh\u01b0 v\u1eady: $\\widehat{DIF}=60^o$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 $\\widehat{DIE}=60^o;\\widehat{DIF}=60^o$ <\/span> <\/span> "}]}],"id_ques":1485},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t <br\/> Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" Cho tam gi\u00e1c $ABC$ \u0111\u1ec1u. G\u1ecdi $M\\in BC$, $E$ v\u00e0 $F$ l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c k\u1ebb t\u1eeb $M$ \u0111\u1ebfn $AB, AC.$ G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AM, D$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC.$ <br\/> C\u00e2u 2: <\/b> $DEIF$ l\u00e0 h\u00ecnh thoi <\/span>","select":[" \u0110\u00fang"," Sai"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_4.jpg' \/><\/center> Tam gi\u00e1c $DIE$ c\u00e2n t\u1ea1i $I$ ($EI=DI=IA$) <br\/> M\u00e0 theo c\u00e2u 1, ta t\u00ednh \u0111\u01b0\u1ee3c: $\\widehat{DIE}=60^o$ n\u00ean $\\Delta DIE$ \u0111\u1ec1u <br\/> T\u01b0\u01a1ng t\u1ef1, ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c $\\Delta DIF$ \u0111\u1ec1u <br\/> Do \u0111\u00f3: $EI=ED=DF=IF$ <br\/> $\\Rightarrow $ T\u1ee9 gi\u00e1c $DEIF$ l\u00e0 h\u00ecnh thoi. <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang. <\/span><\/span> ","column":2}]}],"id_ques":1486},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho h\u00ecnh thoi $ABCD$ c\u00f3 $\\widehat{A}=60^o$. Tr\u00ean c\u1ea1nh $AD$ l\u1ea5y \u0111i\u1ec3m $M$, tr\u00ean c\u1ea1nh $DC$ l\u1ea5y \u0111i\u1ec3m $N$ sao cho $AM=DN.$ <br\/> C\u00e2u 1: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $BDC$ <br\/><br\/> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{BDC}= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_5a.jpg' \/><\/center> Theo b\u00e0i, h\u00ecnh thoi $ABCD$ c\u00f3 $\\widehat{A}=60^o$ <br\/> M\u00e0 $\\widehat{A}+\\widehat{B}=180^o$ <br\/> $\\Rightarrow \\widehat{B}=\\widehat{D}=120^o$ <br\/> M\u00e0 $DB$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $D$ trong h\u00ecnh thoi $ABCD$ <br\/> $\\Rightarrow \\widehat{BDC}=\\dfrac{\\widehat{D}}{2}=60^o$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 $\\widehat{BDC}=60^o$ <\/span> <\/span> "}]}],"id_ques":1487},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["DBN"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho h\u00ecnh thoi $ABCD$ c\u00f3 $\\widehat{A}=60^o$. Tr\u00ean c\u1ea1nh $AD$ l\u1ea5y \u0111i\u1ec3m $M$, tr\u00ean c\u1ea1nh $DC$ l\u1ea5y \u0111i\u1ec3m $N$ sao cho $AM=DN.$ <br\/> C\u00e2u 2: <\/b> Ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c $\\Delta ABM=\\Delta$ ___________ <br\/><br\/> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\Delta ABM=\\Delta \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_5a.jpg' \/><\/center> X\u00e9t $\\Delta ABD$ c\u00f3:<br\/> $AB = AD$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi) <br\/> $\\widehat{BAD} = 60^o$ (gi\u1ea3 thi\u1ebft) <br\/> Suy ra $\\Delta BAD$ \u0111\u1ec1u <br\/> X\u00e9t hai tam gi\u00e1c $ABM$ v\u00e0 $DBN$ c\u00f3: <br\/> + $AB=BD$ ($\\Delta ABD$ \u0111\u1ec1u) <br\/> + $ \\widehat{BAM}=\\widehat{BDN}=60^o$ <br\/> + $AM=DN$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\Delta ABM=\\Delta DBN$ (c - g - c) <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 $\\Delta ABM=\\Delta DBN$ <\/span> <\/span> "}]}],"id_ques":1488},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho h\u00ecnh thoi $ABCD$ c\u00f3 $\\widehat{A}=60^o$. Tr\u00ean c\u1ea1nh $AD$ l\u1ea5y \u0111i\u1ec3m $M$, tr\u00ean c\u1ea1nh $DC$ l\u1ea5y \u0111i\u1ec3m $N$ sao cho $AM=DN$ <br\/> C\u00e2u 3: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $MBN$. <br\/><br\/> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{MBN}= \\, \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span> ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_5d.jpg' \/><\/center> Ta c\u00f3: $\\widehat{ABM}=\\widehat{DBN}$ (do $\\Delta ABM=\\Delta DBN$) <br\/> M\u00e0 $\\widehat{ABM}+\\widehat{B_1}=60^o$ <br\/> $\\Rightarrow \\widehat{DBN}+\\widehat{B_1}=60^o$ <br\/> $\\Rightarrow \\widehat{MBN}=60^o$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 $\\widehat{MBN}=60^o$ <\/span> <\/span> "}]}],"id_ques":1489},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" Cho h\u00ecnh thoi $ABCD$ c\u00f3 $\\widehat{A}=60^o$. Tr\u00ean c\u1ea1nh $AD$ l\u1ea5y \u0111i\u1ec3m $M$, tr\u00ean c\u1ea1nh $DC$ l\u1ea5y \u0111i\u1ec3m $N$ sao cho $AM=DN$ <br\/> C\u00e2u 4: <\/b> $\\Delta BMN$ l\u00e0 tam gi\u00e1c g\u00ec? <\/span>","select":["A. Tam gi\u00e1c c\u00e2n","B. Tam gi\u00e1c \u0111\u1ec1u","C. Tam gi\u00e1c vu\u00f4ng","D. Tam gi\u00e1c vu\u00f4ng c\u00e2n"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai7/lv3/img\/H817_K1_5d.jpg' \/><\/center> Theo c\u00e2u 2, ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\Delta ABM=\\Delta DBN$ <br\/> $\\Rightarrow BM=BN$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow \\Delta MBN$ c\u00e2n t\u1ea1i $B$. <br\/> M\u00e0 theo c\u00e2u 3: $\\widehat{NBM}=60^o$ <br\/> $\\Rightarrow \\Delta BMN$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u. <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span> ","column":2}]}],"id_ques":1490}],"lesson":{"save":0,"level":3}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên	+ 5 điểm
Trong khoảng 1 phút -> 2 phút	+ 4 điểm
Trong khoảng 2 phút -> 3 phút	+ 3 điểm
Trong khoảng 3 phút -> 4 phút	+ 2 điểm
Trong khoảng 4 phút -> 5 phút	+ 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

Điểm của bạn.

Bấm vào đây nếu phát hiện có lỗi hoặc muốn gửi góp ý

Em chưa làm xong câu này