{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho h\u00e0m s\u1ed1 $y = \\dfrac{-1}{2}x $ c\u00f3 \u0111\u1ed3 th\u1ecb l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d v\u00e0 c\u00e1c \u0111i\u1ec3m $M(0 ; 0), N\\left(\\dfrac{1}{2}; \\dfrac{-1}{4} \\right),$$ G(4; -2), H\\left(\\dfrac{-1}{3}; \\dfrac{-1}{6} \\right), E\\left(\\sqrt{2}; \\dfrac{-\\sqrt{2}}{2} \\right), F(2; 1) . $ Trong c\u00e1c \u0111i\u1ec3m \u0111\u00e3 cho \u0111i\u1ec3m n\u00e0o thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1? ","select":["A. \u0110i\u1ec3m M v\u00e0 N ","B. \u0110i\u1ec3m M v\u00e0 H ","C. \u0110i\u1ec3m E v\u00e0 F ","D. \u0110i\u1ec3m M, N, G, E "],"explain":" <span class='basic_left'> $\\bullet $ Thay $x = 0 $ v\u00e0o $y = \\dfrac{-1}{2}x $ ta \u0111\u01b0\u1ee3c $y = \\dfrac{-1}{2}.0 = 0 $ <br\/> $\\Rightarrow $ \u0111i\u1ec3m $M(0; 0)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{-1}{2}x $ <br\/> $\\bullet $ Thay $x = \\dfrac{1}{2} $ v\u00e0o $y = \\dfrac{-1}{2}x $ ta \u0111\u01b0\u1ee3c $y = \\dfrac{-1}{2}.\\dfrac{1}{2} = \\dfrac{-1}{4} $ <br\/> $\\Rightarrow $ \u0111i\u1ec3m $N\\left(\\dfrac{1}{2}; \\dfrac{-1}{4} \\right)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{-1}{2}x $ <br\/> $\\bullet $ Thay $x = 4 $ v\u00e0o $y = \\dfrac{-1}{2}x $ ta \u0111\u01b0\u1ee3c $y = \\dfrac{-1}{2}.4 = -2 $ <br\/> $\\Rightarrow $ \u0111i\u1ec3m $G(4; -2)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{-1}{2}x $ <br\/> $\\bullet $ Thay $x = \\dfrac{-1}{3} $ v\u00e0o $y = \\dfrac{-1}{2}x $ ta \u0111\u01b0\u1ee3c $y = \\dfrac{-1}{2}.\\dfrac{-1}{3} = \\dfrac{1}{6} \\neq \\dfrac{-1}{6} $ <br\/> $\\Rightarrow $ \u0111i\u1ec3m $H\\left(\\dfrac{-1}{3}; \\dfrac{-1}{6} \\right)$ kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{-1}{2}x $ <br\/> $\\bullet $ Thay $x = \\sqrt{2} $ v\u00e0o $y = \\dfrac{-1}{2}x $ ta \u0111\u01b0\u1ee3c $y = \\dfrac{-1}{2}.\\sqrt{2} = \\dfrac{-\\sqrt{2}}{2} $ <br\/> $\\Rightarrow $ \u0111i\u1ec3m $E\\left(\\sqrt{2}; \\dfrac{-\\sqrt{2}}{2} \\right)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{-1}{2}x $ <br\/> $\\bullet $ Thay $x = 2 $ v\u00e0o $y = \\dfrac{-1}{2}x $ ta \u0111\u01b0\u1ee3c $y = \\dfrac{-1}{2}.2 = -1 \\neq 1 $ <br\/> $\\Rightarrow $ \u0111i\u1ec3m $F(2; 1)$ kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{-1}{2}x $ <br\/> V\u1eady c\u00f3 4 b\u1ed1n \u0111i\u1ec3m $M, N, G, E$ thu\u1ed9c \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho. <br\/><br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 D. <\/span><br\/>","column":2}]}],"id_ques":801},{"time":14,"part":[{"title":"V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u ","temp":"coordinates","correct":[["1,-3","0.5,-1.5","1.5,-4.5","-1,3","-0.5,1.5","-1.5,4.5"]],"list":[{"point":10,"toa_do":[["0","0",""]],"is_click":1,"name_toado_click":"A","draw_line":1,"ques":"Cho h\u00e0m s\u1ed1 $y = (1 - 4a)x $ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(-2; 6).$ <br\/> <u> C\u00e2u 1: <\/u> Vi\u1ebft c\u00f4ng th\u1ee9c v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 n\u00f3i tr\u00ean. <br\/> <br\/>_inputduongthang_ <br\/> 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.","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>- T\u00ecm a \u0111\u1ec3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i qua \u0111i\u1ec3m $A(-2; 6)$ <br\/> - Thay a v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o h\u00e0m s\u1ed1 \u0111\u1ec3 \u0111\u01b0\u1ee3c c\u00f4ng th\u1ee9c c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho. <br\/> - V\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = (1 - 4a)x $ \u0111i qua \u0111i\u1ec3m $A(-2; 6)$ khi: <br\/> $\\begin{aligned} & 6=(1-4a)(-2) \\\\ & \\Leftrightarrow 1 - 4a = 6 : (-2) = -3 \\\\ & \\Leftrightarrow 4a = 1 + 3 = 4 \\\\ & \\Leftrightarrow a = 1 \\\\ \\end{aligned}$ <br\/> V\u1edbi $a = 1 $ ta c\u00f3: $y = (1 - 4.1)x = -3x $ <br\/> V\u1eady c\u00f4ng th\u1ee9c c\u1ee7a h\u00e0m n\u00f3i tr\u00ean l\u00e0 $y = -3x$ <br\/> \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = -3x$ l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua $O(0; 0)$ v\u00e0 $A(1; -3)$ <\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai17/lv3/img\/D7B17-28.png' \/><\/center>"}]}],"id_ques":802},{"time":14,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"checkbox","correct":[["1","2","4"]],"list":[{"point":10,"img":"","ques":"Cho h\u00e0m s\u1ed1 $y = (1 - 4a)x $ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(-2; 6).$ <br\/> <u> C\u00e2u 2: <\/u> Nh\u1eefng \u0111i\u1ec3m n\u00e0o sau \u0111\u00e2y th\u1eb3ng h\u00e0ng? ","column":1,"number_true":3,"select":["$M(1; -3) $","$N\\left(\\dfrac{-1}{3}; 1 \\right) $","$P\\left(\\dfrac{-1}{3}; -1 \\right) $","$Q\\left(\\dfrac{1}{2}; -1,5 \\right) $"],"explain":"\u0110i\u1ec3m M(1; -3) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = -3x $ v\u00ec $-3 = -3.1 $ <br\/> \u0110i\u1ec3m $N\\left(\\dfrac{-1}{3}; 1 \\right) $ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = -3x $ v\u00ec $1 = -3.\\dfrac{-1}{3} $ <br\/> \u0110i\u1ec3m $P\\left(\\dfrac{-1}{3}; -1 \\right) $ kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = -3x $ v\u00ec $-1 \\neq -3.\\dfrac{-1}{3} = 1 $ <br\/> \u0110i\u1ec3m $Q\\left(\\dfrac{1}{2}; -1,5 \\right) $ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = -3x $ v\u00ec $-1,5 = -3.\\dfrac{1}{2} $ <br\/> V\u1eady c\u00f3 3 \u0111i\u1ec3m M, N, Q th\u1eb3ng h\u00e0ng v\u00ec n\u00f3 c\u00f9ng thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = -3x $ "}]}],"id_ques":803},{"time":14,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","temp":"checkbox","correct":[["1","2","3"]],"list":[{"point":10,"img":"","ques":" Trong c\u00e1c h\u00e0m s\u1ed1 sau h\u00e0m s\u1ed1 n\u00e0o c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(-2; 6)?$ ","column":1,"number_true":2,"select":["$y = 3x + 12 $","$y = 2 - 2x $","$y = 3x^{2} - 6 $"],"explain":"H\u00e0m s\u1ed1 $y = 3x + 12 $ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(-2; 6)$ v\u00ec: $6 = 3.(-2) + 12 = 6 $ <br\/> H\u00e0m s\u1ed1 $y =2 - 2x $ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(-2; 6)$ v\u00ec: $ 2 - 2.(-2) = 6 $ <br\/> H\u00e0m s\u1ed1 $y = 3x^{2} - 6 $ c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(-2; 6) $v\u00ec: $6 = 3.(-2)^{2} - 6 = 12 - 6 = 6 $ <br\/> V\u1eady c\u1ea3 ba h\u00e0m s\u1ed1 \u0111\u1ec1u c\u00f3 \u0111\u1ed3 th\u1ecb \u0111i qua \u0111i\u1ec3m $A(-2; 6)$ "}]}],"id_ques":804},{"time":14,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-3"],["2"]]],"list":[{"point":10,"width":50,"ques":" <span class='basic_left'> Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy,$ cho \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = ax$ l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $OA$ v\u1edbi \u0111i\u1ec3m $A(-3; 2)$ <br\/> \u0110i\u1ec3m $B(x_{0}; y_{0}) $ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean. Khi \u0111\u00f3, gi\u00e1 tr\u1ecb c\u1ee7a $\\dfrac{x_{0}-3}{y_{0}+2} = $ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\" style=\"margin-left:10px\">_input_<\/div><\/div><\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>- X\u00e1c \u0111\u1ecbnh c\u00f4ng th\u1ee9c c\u1ee7a h\u00e0m s\u1ed1 n\u00f3i tr\u00ean. <br\/> - T\u00ecm $\\dfrac{x_{0}}{y_{0}} $; t\u1eeb \u0111\u00f3 t\u00ecm \u0111\u01b0\u1ee3c $\\dfrac{x_{0}-3}{y_{0}+2}$. <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>\u0110i\u1ec3m $A(-3; 2)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = ax,$ thay $x = -3, y = 2$ v\u00e0o h\u00e0m s\u1ed1 ta \u0111\u01b0\u1ee3c: <br\/> $2 = -3a \\Rightarrow a = \\dfrac{-2}{3} $ <br\/> C\u00f4ng th\u1ee9c c\u1ee7a h\u00e0m s\u1ed1 n\u00f3i tr\u00ean l\u00e0: $y = \\dfrac{-2}{3}x $ <br\/> \u0110i\u1ec3m $B(x_{0}; y_{0}) $ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{-2}{3}x $ n\u00ean $y_{0} = \\dfrac{-2}{3}x_{0} $ <br\/> $\\Rightarrow \\dfrac{{{x}_{0}}}{{{y}_{0}}}=\\dfrac{-3}{2}$ <br\/> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: <br\/> $\\dfrac{{{x}_{0}}}{{{y}_{0}}}=\\dfrac{-3}{2}=\\dfrac{{{x}_{0}}-3}{{{y}_{0}}+2}$<\/span>"}]}],"id_ques":805},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy,$ cho \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{3}{2}x - \\dfrac{5}{2} .$ <br\/> T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m A, bi\u1ebft A l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 v\u1edbi tr\u1ee5c ho\u00e0nh. ","select":["A. $A\\left(\\dfrac{5}{3}; 0 \\right) $","B. $A\\left(0; \\dfrac{5}{3}\\right) $","C. $A\\left(0; \\dfrac{-5}{2} \\right) $ ","D. $A\\left(\\dfrac{-5}{2}; 0 \\right) $ "],"explain":" A l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 v\u1edbi tr\u1ee5c ho\u00e0nh. <br\/> A thu\u1ed9c tr\u1ee5c ho\u00e0nh $\\Rightarrow $ T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m A c\u00f3 d\u1ea1ng $A(x_{A}; 0)$ <br\/> A thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\dfrac{3}{2}x - \\dfrac{5}{2} $ n\u00ean $0 = \\dfrac{3}{2}.x_{A} - \\dfrac{5}{2} \\\\ \\Rightarrow x_{A} = \\dfrac{5}{3} $ <br\/> V\u1eady t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m A l\u00e0 $A\\left(\\dfrac{5}{3}; 0\\right) $ <br\/><br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span><br\/>","column":2}]}],"id_ques":806},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" X\u00e1c \u0111\u1ecbnh m, bi\u1ebft r\u1eb1ng \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y = \\left|m - \\dfrac{3}{4} \\right|x + 2 $ \u0111i qua \u0111i\u1ec3m M(2; 3) ","select":["A. $m = \\dfrac{5}{4} $","B. $m = \\dfrac{-1}{4} $","C. $m = \\dfrac{5}{4} $ ho\u1eb7c $m = \\dfrac{1}{4}$ ","D. $m = \\dfrac{5}{4} $ ho\u1eb7c $m = \\dfrac{-1}{4}$ "],"explain":" \u0110i\u1ec3m M(2; 3) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = \\left|m - \\dfrac{3}{4} \\right|x + 2 $ n\u00ean ta c\u00f3: <br\/> $\\begin{aligned}& 3=\\left| m-\\dfrac{3}{4} \\right|.2+2 \\\\ \\\\ & \\Leftrightarrow \\left| m-\\dfrac{3}{4} \\right|=\\dfrac{1}{2} \\\\ \\\\ & \\Rightarrow \\left[ \\begin{aligned} & m-\\dfrac{3}{4}=\\dfrac{1}{2} \\\\ & m-\\dfrac{3}{4}=\\dfrac{-1}{2} \\\\ \\end{aligned} \\right. \\\\ \\\\ & \\Rightarrow \\left[ \\begin{aligned}& m=\\dfrac{5}{4} \\\\ & m=\\dfrac{1}{4} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span><br\/>","column":2}]}],"id_ques":807},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" Cho h\u00e0m s\u1ed1 $y = f(x) = ax + b $ <br\/> X\u00e1c \u0111\u1ecbnh c\u00e1c gi\u00e1 tr\u1ecb a, b bi\u1ebft \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 f(x) \u0111i qua hai \u0111i\u1ec3m $M(0; 1) $ v\u00e0 $N\\left(\\dfrac{-1}{2}; 0 \\right)$ ","select":["A. $a = 2; b = -1 $","B. $a = 2; b = 1 $ ","C. $a = -1; b = 1 $ ","D. $a = 1; b = 2 $ "],"explain":" <span class='basic_left'> \u0110i\u1ec3m $M(0; 1) $ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = f(x) = ax + b $ n\u00ean ta c\u00f3: <br\/> $1 = a.0 + b \\Rightarrow b = 1 $ <br\/> \u0110i\u1ec3m $N\\left(\\dfrac{-1}{2}; 0 \\right)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = f(x) = ax + b $ n\u00ean ta c\u00f3: <br\/> $0 = a.\\dfrac{-1}{2} + b \\\\ \\Leftrightarrow 0 = \\dfrac{-1}{2}a + 1 \\\\ \\Rightarrow a = 2 $ <br\/> V\u1eady a = 2, b = 1. <br\/><br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span><br\/>","column":2}]}],"id_ques":808},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"],["1"],["1"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"Cho h\u00e0m s\u1ed1 $y = ax^{2} + bx + c $. <br\/> T\u00ecm a, b, c bi\u1ebft \u0111i\u1ec3m $A(0; 1), B(1; 2), C(-1; 0)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $\\left\\{ \\begin{aligned}& a = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & b = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & c = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$ ","explain":"\u0110i\u1ec3m $A(0; 1), B(1; 2), C(-1; 0)$ thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = ax^{2} + bx + c $ n\u00ean ta c\u00f3: <br\/> $\\begin{aligned} & \\left\\{\\begin{aligned} & 1=a{{.0}^{2}}+b.0+c \\\\ & 2=a{{.1}^{2}}+b.1+c \\\\ & 0=a.{{(-1)}^{2}}+b.(-1)+c \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & 1=c \\\\ & 2=a+b+c \\\\ & 0=a-b+c \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & c=1 \\\\ & a+b=1 \\\\ & a-b=-1 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & c=1 \\\\ & a=0 \\\\ & b=1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $0; 1; 1$ <\/span> "}]}],"id_ques":809},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" H\u00ecnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y = 2|x| $ <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai17/lv3/img\/D7B17-30A.png' \/><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai17/lv3/img\/D7B17-30B.png' \/><\/center> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai17/lv3/img\/D7B17-30C.png' \/><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai17/lv3/img\/D7B17-30D.png' \/><\/center> ","select":["A. H\u00ecnh 1 ","B. H\u00ecnh 2 ","C. H\u00ecnh 3 ","D. H\u00ecnh 4 "],"explain":" Ta c\u00f3: $y = 2|x| = \\begin{cases}2x & n\u1ebfu \\quad x \\geq 0 \\\\ -2x & n\u1ebfu \\quad x < 0\\end{cases}$ <br\/><br\/> B\u1ea3ng gi\u00e1 tr\u1ecb: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai17/lv3/img\/D7B17-30.png' \/><\/center> \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y = 2|x| $ l\u00e0 hai tia OA, OB v\u1edbi $A(1; 2)$ v\u00e0 $B(-1; 2)$ <br\/> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai17/lv3/img\/D7B17-30A.png' \/><\/center> <br\/><br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span><br\/>","column":2}]}],"id_ques":810}],"lesson":{"save":0,"level":3}}