{"segment":[{"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang ","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"H\u00e0m s\u1ed1 $y= (2m-6)x+3-2m \\,\\,(m\\ne 3)$ ngh\u1ecbch bi\u1ebfn khi:","select":["A. $m=3$ ","B. $m < 3$","C. $m > 3$ ","D. $m \\ge 3$"],"hint":"H\u00e0m s\u1ed1 $y = ax +b \\,\\,(a\\ne 0)$ \u0111\u1ed3ng bi\u1ebfn n\u1ebfu $a > 0$ v\u00e0 ngh\u1ecbch bi\u1ebfn n\u1ebfu $a < 0$. ","explain":" H\u00e0m s\u1ed1 $y= (2m-6)x+3-2m \\,\\,(m\\ne 3)$ ngh\u1ecbch bi\u1ebfn khi: <br\/>$2m - 6 < 0 \\Leftrightarrow m < 3$<br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 B.<\/span>","column":4}],"id_ques":201},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"\u0110\u01b0\u1eddng th\u1eb3ng $(m-1)x+(m+1)y=3$ song song v\u1edbi tr\u1ee5c ho\u00e0nh c\u1ee7a h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 $Oxy$ khi $m=$ _input_","hint":"Hai \u0111\u01b0\u1eddng th\u1eb3ng song song khi h\u1ec7 s\u1ed1 g\u00f3c b\u1eb1ng nhau v\u00e0 tung \u0111\u1ed9 g\u1ed1c kh\u00e1c nhau.","explain":"<span class='basic_left'> Tr\u1ee5c ho\u00e0nh c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $y= 0$ n\u00ean h\u1ec7 s\u1ed1 g\u00f3c b\u1eb1ng $0$. <br\/>\u0110\u01b0\u1eddng th\u1eb3ng $(m-1)x+(m+1)y=3$ hay $y=\\dfrac{1-m}{m+1}x+\\dfrac{3}{m+1}$ song song v\u1edbi tr\u1ee5c ho\u00e0nh c\u1ee7a h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 $Oxy$ <br\/>$\\Leftrightarrow 1-m = 0\\Leftrightarrow m = 1$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$.<\/span>"}],"id_ques":202},{"title":"V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y= ax + b$","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"coordinates","correct":[["0,-0.5","1,-2","2,-3.5","3,5","-1,1","-2,2.5"]],"list":[{"point":10,"toa_do":[["-3","4","A"]],"is_click":1,"name_toado_click":"B","draw_line":1,"ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, (m - 2)x+(m - 1)y = 1$<br\/><b> C\u00e2u a: <\/b> V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 v\u1edbi $m = -1$. <br\/>Cho s\u1eb5n $A(-3; 4)$. H\u00e3y click th\u00eam m\u1ed9t \u0111i\u1ec3m thu\u1ed9c \u0111\u1ed3 th\u1ecb \u0111\u1ec3 v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.<br\/>_inputduongthang_ ","explain":" <span class='basic_left'> V\u1edbi $m=- 1 \\,\\Rightarrow -3x-2y=1\\,$$ \\Leftrightarrow y =-\\dfrac{3}{2}x-\\dfrac{1}{2}$<br\/>V\u1edbi $x = 0 \\, \\Rightarrow y =-\\dfrac{1}{2}\\Rightarrow B \\left( 0; -\\dfrac{1}{2}\\right)$<br\/> \u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m $A;B$ l\u00e0 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y =-\\dfrac{3}{2}x-\\dfrac{1}{2}$. <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai13/lv3/img\/D925_K2a.png' \/><\/center>"}],"id_ques":203},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["5"],["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, (m - 2)x+(m - 1)y = 1$<br\/><b> C\u00e2u b: <\/b> T\u00ecm $m$ \u0111\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ \u0111i qua \u0111i\u1ec3m $A (1; 2)$.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b>$m=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$","hint":"Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ \u0111\u1ec3 t\u00ecm $m$.","explain":"Ta c\u00f3: $A(1; 2) \\in (d)$ <br\/> $\\begin {aligned}& \\Rightarrow (m -2).1 +(m-1).2=1\\\\ & \\Leftrightarrow m- 2 + 2m -2=1\\\\&\\Leftrightarrow m=\\dfrac{5}{3} \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $5$ v\u00e0 $3$.<\/span>"}],"id_ques":204},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["-1"],["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, (m - 2)x+(m - 1)y = 1$<br\/><b> C\u00e2u c: <\/b> T\u00ecm \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh $I$ m\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ lu\u00f4n \u0111i qua.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b>T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $I$ l\u00e0 (_input_ ;_input_)","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: G\u1ecdi $I(x_{o},y_{o})$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh thu\u1ed9c $(d)$. Thay t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $I$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $(d)$<br\/>B\u01b0\u1edbc 2: Bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $a.m + b= 0$<br\/>B\u01b0\u1edbc 3: T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ r\u1ed3i k\u1ebft lu\u1eadn. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$(d):\\,(m-2)x+(m-1)y=1$<br\/>G\u1ecdi $I \\left( {{x}_{o}};{{y}_{o}} \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh thu\u1ed9c $(d)$<br\/>$\\Rightarrow I\\left( {{x}_{o}};{{y}_{o}} \\right)\\in \\left( d \\right)\\,\\,\\,\\forall m$ <br\/>$ \\left( m-2 \\right){{x}_{o}}+\\left( m-1 \\right){{y}_{o}}=1\\,\\,\\,\\forall m$ <br\/>$ \\Leftrightarrow m{{x}_{o}}-2x_{o}+m{{y}_{o}}-{{y}_{o}}-1=0\\,\\forall m $<br\/>$ \\Leftrightarrow \\left( {{x}_{o}}+{{y}_{o}} \\right)m-\\left( 2{{x}_{o}}+{{y}_{o}}+1 \\right)=0\\,\\forall m$ <br\/>$ \\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}+y_{o}=0 \\\\ & 2x_{o}+y_{o}+1=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}=-1 \\\\ & {{y}_{o}}=1 \\\\ \\end{aligned} \\right.$ <br\/>$\\Rightarrow I\\left( -1;1 \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh thu\u1ed9c $\\left( d \\right)$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-1;1 $<\/span><\/span>"}],"id_ques":205},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"\u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m $A (4; -6)$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $y=x - 5$ c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $y = -x +3$ ","B. $y = x -8$","C. $y= -2x +2$ ","D. $y= -x -2$"],"hint":"$(d):\\,y= ax + b\\,\\,(a\\ne 0)$ v\u00e0 $(d'):\\, y= a'x+b' \\,\\,(a'\\ne 0)$<br\/>$(d)\\bot (d') \\Leftrightarrow a.a'=-1$ ","explain":"<span class='basic_left'>G\u1ecdi $(d):\\,y=ax+b$. <br\/>\u0110\u01b0\u1eddng th\u1eb3ng $(d)$ vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $y= x - 5$, ta c\u00f3: $a.1=-1\\Rightarrow a=-1$<br\/>Suy ra \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ c\u00f3 d\u1ea1ng: $y=-x+b$<br\/>Do $A(4; -6)$ thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ n\u00ean ta c\u00f3: <br\/>$ -6=-4+ b \\Leftrightarrow b= -2$<br\/>V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng c\u1ea7n t\u00ecm c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $y = -x -2$. <br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 D.<\/span> <\/span>","column":2}],"id_ques":206},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho hai \u0111\u01b0\u1eddng th\u1eb3ng : $y=2x$ v\u00e0 $y=-3x+5$. T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m $M$ c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng $y=2x$ v\u00e0 $y=-3x+5$<br\/><b>\u0110\u00e1p \u00e1n:<\/b> T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $M$ l\u00e0 (_input_; _input_)<\/span>","hint":"X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m.","explain":" <span class='basic_left'> Ta c\u00f3 $(d):\\,\\,y=2x\\,\\,$ v\u00e0 $(d'): \\,\\,y=-3x+5$<br\/> Theo b\u00e0i: $(d) \\cap (d') =M$. <br\/> Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m $M$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$2x=-3x+5\\Leftrightarrow x=1$<br\/>Tung \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $M$ l\u00e0 $y= 2$.<br\/>V\u1eady t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $M$ l\u00e0 $(1; 2)$. <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec3n l\u00e0 $1;2$<\/span>"}],"id_ques":207},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"\u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m $A (2; 5)$ v\u00e0 $B (1; -3)$ c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $y=8x - 11$","B. $y=8x + 11$","C. $y=8x -5$ "],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ c\u00f3 d\u1ea1ng: $y = ax + b\\, (a\\ne 0)$<br\/>Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$ v\u00e0 $B$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ec3 t\u00ecm $a;b$. ","explain":"<span class='basic_left'> Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ c\u00f3 d\u1ea1ng $y= ax + b\\, (a \\ne 0)$.<br\/>Do $A\\left( 2;5 \\right)\\,\\,\\in AB$ n\u00ean t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$ th\u1ecfa m\u00e3n ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$. <br\/> $\\Rightarrow 5=a.2+b\\,$$\\Leftrightarrow 2a+b=5\\,\\,\\,\\,(1) $<br\/>T\u01b0\u01a1ng t\u1ef1: $B\\left( 1;-3 \\right)\\in AB$$\\Rightarrow -3=a.1+b\\,$$\\Leftrightarrow b=-a-3\\,\\,\\,\\,(2) $<br\/>Thay (2) v\u00e0o (1) ta \u0111\u01b0\u1ee3c: $2a-a-3=5\\Leftrightarrow a=8 $<br\/> Thay $a=8$ v\u00e0o (2) \u0111\u01b0\u1ee3c: $b= -8-3=-11$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ l\u00e0 : $y=8x-11$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":3}],"id_ques":208},{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{60}{13}$","B. $\\dfrac{55}{13}$","C. $\\dfrac{53}{13}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai13/lv3/img\/\/2.png' \/><\/center>Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,\\,\\,y=-\\dfrac{12}{5}x+12$. <br\/>Kho\u1ea3ng c\u00e1ch t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 $O$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ l\u00e0 ?","hint":"D\u00f9ng h\u1ec7 th\u1ee9c gi\u1eefa \u0111\u01b0\u1eddng cao v\u00e0 c\u1ea1nh trong tam gi\u00e1c vu\u00f4ng $AOB$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1:X\u00e1c \u0111\u1ecbnh giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb v\u1edbi hai tr\u1ee5c t\u1ecda \u0111\u1ed9. <br\/>B\u01b0\u1edbc 2: D\u00f9ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3 $\\left( d \\right)\\cap \\text{Ox}=A(5;0);\\,\\,$$\\left( d \\right)\\cap Oy=B\\left( 0;12 \\right)$. H\u1ea1 $OH\\bot AB$ . <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai13/lv3/img\/\/D925_K1.png' \/><\/center><br\/> Theo h\u00ecnh v\u1ebd ta x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c: $OA=5;OB=12$. <br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng $AOB$ ta c\u00f3: <br\/>$\\begin{aligned}\\dfrac{1}{O{{H}^{2}}}&=\\dfrac{1}{O{{A}^{2}}}+\\dfrac{1}{O{{B}^{2}}}\\\\&=\\dfrac{1}{{{5}^{2}}}+\\dfrac{1}{{{12}^{2}}}=\\dfrac{169}{3600}\\\\&\\Rightarrow OH=\\dfrac{60}{13}\\end{aligned}$ <br\/>Kho\u1ea3ng c\u00e1ch t\u1eeb O \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $\\left( d \\right)$ l\u00e0 $\\dfrac{60}{13}$<br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> \u0110\u1ec3 gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp d\u1ea1ng t\u00ecm kho\u1ea3ng c\u00e1ch t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng, ta th\u01b0\u1eddng s\u1eed d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng:<br\/>+ $\\dfrac{1}{O{{H}^{2}}}=\\dfrac{1}{O{{A}^{2}}}+\\dfrac{1}{O{{B}^{2}}}$<br\/>+ $OH.AB=OA.OB$ <span>"}],"id_ques":209},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"T\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 $y=\\sqrt{4-3x}+\\dfrac{1}{\\sqrt{x}}$ l\u00e0: ","select":["A. $x \\ge 0$ ","B. $x \\ge \\dfrac{4}{3}$","C. $x \\le \\dfrac{4}{3}$","D. $0 < x \\le \\dfrac{4}{3}$"],"hint":"Bi\u1ec3u th\u1ee9c x\u00e1c \u0111\u1ecbnh khi bi\u1ec3u th\u1ee9c trong c\u0103n kh\u00f4ng \u00e2m v\u00e0 m\u1eabu th\u1ee9c kh\u00e1c $0$. ","explain":"<span class='basic_left'> T\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 $y=\\sqrt{4-3x}+\\dfrac{1}{\\sqrt{x}}$ l\u00e0: <br\/>$\\left\\{ \\begin{aligned} & 4-3x\\ge 0 \\\\ & x > 0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\le \\dfrac{4}{3} \\\\ & x > 0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow 0< x\\le \\dfrac{4}{3}$ <br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}],"id_ques":210}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}