{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["-1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=-x+2$ v\u00e0 $y=2x-7$. <br\/>T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m $N$ c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng $y=-x+2\\,$ v\u00e0 $y=2x-7$<br\/><br\/><b>\u0110\u00e1p \u00e1n:<\/b> T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $N$ 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 $(d):\\,\\,y=-x+2$ v\u00e0 $(d'): \\,\\,y=2x-7$<br\/>$(d) \\cap (d')=N$. X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a $(d)$ v\u00e0 $(d')$:<br\/>$-x+2=2x-7\\Leftrightarrow x=3$<br\/>Thay $x=3$ v\u00e0o $y=-x+2$ \u0111\u01b0\u1ee3c: $y=-3+2 \\Leftrightarrow y=-1$ <br\/> V\u1eady t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $N(3; -1)$. <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3;-1$.<\/span>"}]}],"id_ques":91},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-3"],["7"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/2.jpg' \/><\/center> <span class='basic_left'> Cho h\u00e0m s\u1ed1 $y= (m-2)x+3m+1$. T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh $I$ m\u00e0 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 lu\u00f4n \u0111i qua v\u1edbi m\u1ecdi $m$.<br\/>T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $I$ l\u00e0 (_input_ ; _input_) <\/span>","hint":"G\u1ecdi $I(x_0,y_0)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh c\u1ea7n t\u00ecm. Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $I$ v\u00e0o h\u00e0m s\u1ed1. \u0110\u01b0a b\u00e0i to\u00e1n v\u1ec1 t\u00ecm $x_0,y_0$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi $m$. ","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: \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh (d) v\u1ec1 d\u1ea1ng $am+b=0$ v\u1edbi m\u1ecdi $m$. Khi \u0111\u00f3 $a=0$ v\u00e0 $b=0$. <br\/>B\u01b0\u1edbc 3: Suy ra $x_0,y_0$ v\u00e0 k\u1ebft lu\u1eadn <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>G\u1ecdi \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y= (m-2)x+3m+1$ l\u00e0 $(d)$<br\/>G\u1ecdi $I \\left( {{x}_{o}};{{y}_{o}} \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh c\u1ee7a $(d)$<br\/>$\\Rightarrow I\\left( {{x}_{o}};{{y}_{o}} \\right)\\in \\left( d \\right)\\,\\,\\,\\forall m$ <br\/>$\\begin{aligned} & \\Leftrightarrow {{y}_{o}}=\\left( m-2 \\right){{x}_{o}}+3m+1\\,\\,\\,\\forall m \\\\ & \\Leftrightarrow m{{x}_{o}}-2{{x}_{o}}+3m+1-{{y}_{o}}\\,\\,\\,\\,=0\\,\\,\\,\\,\\forall m \\\\ & \\Leftrightarrow \\left( {{x}_{o}}+3 \\right)m-2{{x}_{o}}-{{y}_{o}}+1=0\\,\\,\\,\\,\\forall m \\\\ \\end{aligned}$<br\/>$ \\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}+3=0 \\\\ & -2{{x}_{o}}-{{y}_{o}}+1=0 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}=-3 \\\\ & {{y}_{o}}=(-2).(-3)+1=7 \\\\ \\end{aligned} \\right. $ <br\/>$\\Rightarrow I\\left( -3;7 \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh $\\in \\left( d \\right)$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-3;7$<\/span><\/span>"}]}],"id_ques":92},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"\u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y=|x-1|$ l\u00e0:","select":["A. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/D922_K3a.png' \/><\/center>","B. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/D922_K3b.png' \/><\/center>"],"hint":"$|x|=\\left\\{ \\begin{align} & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}\\ge \\text{0} \\\\ & -x\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}< \\text{0} \\\\\\end{align} \\right.$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Ph\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i c\u1ee7a h\u00e0m s\u1ed1 $y$.<br\/> B\u01b0\u1edbc 2: V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u1ee9ng v\u1edbi t\u1eebng kho\u1ea3ng x\u00e1c \u0111\u1ecbnh \u1edf b\u01b0\u1edbc 1. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> V\u1edbi $x-1\\ge 0\\Rightarrow x \\ge 1\\,\\,$ th\u00ec $y=x-1$ <br\/>V\u1edbi $x -1< 0 \\Rightarrow x <1\\,\\,$ th\u00ec $y=-x+1$ <br\/>V\u1eady $y=\\left\\{ \\begin{align} & x-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}\\ge \\text{1} \\\\ & -x+1\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}< \\text{1} \\\\\\end{align} \\right.$<br\/>V\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng $y = x-1$ v\u00e0 $y =-x+1$ tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 . \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0 \u0111\u01b0\u1eddng g\u1ea5p kh\u00fac li\u1ec1n n\u00e9t \u1edf h\u00ecnh d\u01b0\u1edbi.<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/D922_K3a.png' \/><\/center><br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> V\u00ec $y=|x-1|\\ge 0$ n\u00ean ta c\u00f3 th\u1ec3 v\u1ebd \u0111\u1ed3 th\u1ecb hai h\u00e0m s\u1ed1 $y=x-1$ v\u00e0 $y=-x+1$ r\u1ed3i x\u00f3a ph\u1ea7n \u0111\u1ed3 th\u1ecb ph\u00eda d\u01b0\u1edbi tr\u1ee5c $Ox$<\/span>","column":2}]}],"id_ques":93},{"time":24,"part":[{"title":"V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{2}{5}x- \\dfrac{3}{5}$","title_trans":"","temp":"coordinates","correct":[["1.5,0","-1,-1","-3.5,-2"]],"list":[{"point":10,"toa_do":[["4","1","A"]],"is_click":1,"name_toado_click":"B","draw_line":1,"ques":"Cho s\u1eb5n \u0111i\u1ec3m $A (4; 1)$. 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_ <br\/> ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span> B\u01b0\u1edbc 1: Cho $y=0$ ho\u1eb7c $x=0$ \u0111\u1ec3 t\u00ecm giao \u0111i\u1ec3m $B$ c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng v\u1edbi tr\u1ee5c $Ox$. <br\/>B\u01b0\u1edbc 2: V\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua $A, B$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Cho $y= 0 \\Rightarrow x= \\dfrac{3}{2}\\Rightarrow B \\left( 0;\\dfrac{3}{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{2}{5}x- \\dfrac{3}{5}$.<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/D922_K6a.png' \/><\/center>"}]}],"id_ques":94},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{12}{5}$","B. $\\dfrac{5}{12}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/9.jpg' \/><\/center>Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,\\,\\,y=-\\dfrac{3}{4}x+3\\,\\,$. Kho\u1ea3ng c\u00e1ch t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 $O$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ ?","hint":"","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\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/D922_K8.png' \/><\/center><br\/>Ta c\u00f3:<br\/> $\\left( d \\right)\\cap \\text{Ox}=A(4;0)\\Rightarrow OA=4;\\,\\,$<br\/>$\\left( d \\right)\\cap Oy=B\\left( 0;3 \\right)\\Rightarrow OB=3$.<br\/> H\u1ea1 $OH\\bot AB$. <br\/>X\u00e9t tam gi\u00e1c vu\u00f4ng $AOB$ c\u00f3 $OH\\bot AB$, ta c\u00f3: <br\/>$\\begin{aligned}\\dfrac{1}{O{{H}^{2}}}&=\\dfrac{1}{O{{A}^{2}}}+\\dfrac{1}{O{{B}^{2}}}\\\\&=\\dfrac{1}{{{4}^{2}}}+\\dfrac{1}{{{3}^{2}}}=\\dfrac{25}{144}\\\\&\\Rightarrow OH=\\dfrac{12}{5}\\\\ \\end{aligned}$ <br\/>Kho\u1ea3ng c\u00e1ch t\u1eeb O \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $\\left( d \\right)$ l\u00e0 $\\dfrac{12}{5}$<br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0: $\\dfrac{12}{5}$.<\/span><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":95},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00e0m s\u1ed1 $y=f(x)=|x-1|+|x-4|.$ T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $f(x)$<br\/><br\/>\u0110\u00e1p s\u1ed1: Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $f(x)$ l\u00e0 _input_","hint":"$|a|+|b| \\ge|a+b|$<br\/> D\u1ea5u b\u1eb1ng x\u1ea3y ra khi $a.b \\ge 0$","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{aligned} f\\left( x \\right) & =\\left| x-1 \\right|+\\left| x-4 \\right| \\\\ & =\\left| x-1 \\right|+\\left| 4-x \\right|\\\\&\\ge \\left| x-1+4-x \\right|=3 \\\\ \\end{aligned}$ <br\/>$\\Rightarrow f\\left( x \\right)\\ge 3\\,\\,\\,\\,\\,\\forall x\\in \\mathbb R$ <br\/>D\u1ea5u b\u1eb1ng x\u1ea3y ra $\\Leftrightarrow \\left( x-1 \\right)\\left( 4-x \\right)\\ge 0\\Leftrightarrow 1\\le x\\le 4$<br\/>V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 $f(x)$ l\u00e0 $3$ khi $1\\le x\\le 4$. <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3$. <\/span><br\/><span class='basic_green'><b>Ghi nh\u1edb: <\/b><\/span> B\u1ea5t \u0111\u1eb3ng th\u1ee9c d\u1ea5u gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i:<br\/>+ $|a|-|b|\\le|a+b|\\le|a|+|b|$ D\u1ea5u $=$ x\u1ea9y ra khi $a.b\\ge0$<\/span> <\/span>"}]}],"id_ques":96},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u ","temp":"fill_the_blank","correct":[[["4"],["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho c\u00e1c h\u00e0m s\u1ed1 sau: <br\/>$ \\left( {{d}_{1}} \\right):\\,\\,y=-x+5 \\,\\,\\,;\\,\\,\\,$$\\left( {{d}_{2}} \\right):\\,\\,y=4x $$\\,\\,\\,;\\,\\,\\,$$ \\left( {{d}_{3}} \\right):\\,\\,y=\\dfrac{1}{4}x. $ <br\/><b> C\u00e2u 1: <\/b> Bi\u1ebft $\\,\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{3}} \\right)=A$ . T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $A$ l\u00e0 (_input_; _input_)<\/span>","hint":"X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a $(d_1)$ v\u00e0 $(d_3)$","explain":"<span class='basic_left'> $\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{3}} \\right)=A$. <br\/> X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a $(d_1)$ v\u00e0 $(d_3)$, ta c\u00f3:<br\/>$-x+5=\\dfrac{1}{4}x\\Leftrightarrow -4x+20=x\\Leftrightarrow x=4$ <br\/> Thay $x=4$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $y=-x+5$, ta \u0111\u01b0\u1ee3c: $y=-4+5=1 \\Rightarrow y=1$ <br\/>V\u1eady t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A(4; 1)$. <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $ 4;1$<\/span><\/span>"}]}],"id_ques":97},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["1"],["4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho c\u00e1c h\u00e0m s\u1ed1 sau: <br\/>$ \\left( {{d}_{1}} \\right):\\,\\,y=-x+5 \\,\\,\\,;\\,\\,\\,$$\\left( {{d}_{2}} \\right):\\,\\,y=4x $$\\,\\,\\,;\\,\\,\\,$$ \\left( {{d}_{3}} \\right):\\,\\,y=\\dfrac{1}{4}x. $ <br\/><b> C\u00e2u 2: <\/b> Bi\u1ebft $\\,\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{2}} \\right)=B$ . T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $B$ l\u00e0 (_input_; _input_)<\/span>","hint":"X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a $(d_1)$ v\u00e0 $(d_2)$","explain":"<span class='basic_left'>$\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{2}} \\right)=B$. <br\/> X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a $(d_1)$ v\u00e0 $(d_2)$, ta c\u00f3:<br\/>$-x+5=4x\\Leftrightarrow -5x+5=0\\Leftrightarrow x=1$ <br\/> Thay $x=1$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $y=-x+5$ ta \u0111\u01b0\u1ee3c: $y=-1+5 \\Rightarrow y=4$<br\/>V\u1eady t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $B(1; 4)$ <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1;4$<\/span><\/span>"}]}],"id_ques":98},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho c\u00e1c h\u00e0m s\u1ed1 sau: <br\/>$ \\left( {{d}_{1}} \\right):\\,\\,y=-x+5 \\,\\,\\,;\\,\\,\\,$$\\left( {{d}_{2}} \\right):\\,\\,y=4x $$\\,\\,\\,;\\,\\,\\,$$ \\left( {{d}_{3}} \\right):\\,\\,y=\\dfrac{1}{4}x.$ <br\/><b> C\u00e2u 3: <\/b> Bi\u1ebft $\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{3}} \\right)=A,\\,\\,\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{2}} \\right)=B$. <br\/> $\\Delta OAB$ 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"],"hint":"So s\u00e1nh c\u1ea1nh $OA$ v\u00e0 $OB$ ","explain":"<span class='basic_left'> G\u1ecdi $D$ v\u00e0 $E$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A$ tr\u00ean $Ox$ v\u00e0 $B$ tr\u00ean $Oy$. <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/D922_K11.png' \/><\/center><br\/>X\u00e9t $\\Delta EBO$ v\u00e0 $\\Delta DAO$ <br\/>$\\begin{aligned} & AD=BE=1 \\\\ & \\widehat{BEO}=\\widehat{ADO}={{90}^{0}} \\\\ & OD=OE=4 \\\\ \\end{aligned}$ <br\/>$\\Rightarrow \\Delta EBO=\\Delta DAO\\left( c.g.c \\right)$<br\/> $\\Rightarrow AO=OB$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng)<br\/>$\\Rightarrow \\Delta OAB$ c\u00e2n. <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":3}]}],"id_ques":99},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{15}{2}$","B. $\\dfrac{15}{4}$","C. $\\dfrac{15}{7}$"],"ques":"<span class='basic_left'>Cho c\u00e1c h\u00e0m s\u1ed1 sau: <br\/>$ \\left( {{d}_{1}} \\right):\\,\\,y=-x+5 \\,\\,\\,;\\,\\,\\,$$\\left( {{d}_{2}} \\right):\\,\\,y=4x $$\\,\\,\\,;\\,\\,\\,$$ \\left( {{d}_{3}} \\right):\\,\\,y=\\dfrac{1}{4}x.$ <br\/>Bi\u1ebft $\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{3}} \\right)=A,\\,\\,\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{2}} \\right)=B$. <br\/> Di\u1ec7n t\u00edch $\\Delta OAB$ l\u00e0 ?<\/span>","hint":"$\\left( {{d}_{1}} \\right)\\cap \\text{Ox=N;}\\,\\,\\left( {{d}_{1}} \\right)\\cap Oy=M$. <br\/>$S_{\\Delta OAB}=S_{\\Delta OMN}-S_{\\Delta OMB}-S_{\\Delta ONA}$","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai10/lv3/img\/D922_K11.png' \/><\/center> <br\/>Ta c\u00f3:<br\/>$\\left( {{d}_{1}} \\right)\\cap Oy=M(0;5)\\Rightarrow OM=5$<br\/>$\\left( {{d}_{1}} \\right)\\cap Ox=N(5;0)\\Rightarrow ON=5\\,$<br\/>$\\Rightarrow OM=ON=5\\,$<br\/>$\\begin{aligned} &{{S}_{\\Delta MON}}=\\dfrac{1}{2}ON.OM=\\dfrac{1}{2}.5.5=\\dfrac{25}{2}\\\\&{{S}_{\\Delta MBO}}=\\dfrac{1}{2}BE.OM=\\dfrac{1}{2}.1.5=\\dfrac{5}{2}\\\\&{{S}_{\\Delta AON}}=\\dfrac{1}{2}AD.ON=\\dfrac{1}{2}.1.5=\\dfrac{5}{2} \\\\ \\end{aligned}$ <br\/>$ \\Rightarrow {{S}_{\\Delta AOB}}={{S}_{\\Delta MON}}-{{S}_{\\Delta AON}}-{{S}_{\\Delta MBO}}=$$\\dfrac{25}{2}-\\dfrac{5}{2}-\\dfrac{5}{2}=$$\\dfrac{25}{2}-5=\\dfrac{15}{2}$<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 $\\dfrac{15}{2}$.<\/span><\/span>"}]}],"id_ques":100}],"lesson":{"save":0,"level":3}}