{"segment":[{"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":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai4/lv3/img\/2.jpg' \/><\/center> V\u1edbi $x > 0, y < 0$ v\u00e0 $|x| = |y|$ th\u00ec: ","select":["A. $ x^{2}y > 0$","B. $ x + y = 0 $","C. $ x - y = 0 $","D. $ x = \\pm y $ "],"hint":" $|x| = |y| \\Leftrightarrow x = \\pm y $ ","explain":" Ta c\u00f3 : $|x| = |y|$ <br\/> Suy ra $x = y $ ho\u1eb7c $x = -y $ (1) <br\/> M\u00e0 theo gi\u1ea3 thi\u1ebft c\u00f3 $x > 0, y < 0$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $x = -y $ <br\/> Hay $x + y = 0 $ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> <br\/>","column":2}]}],"id_ques":191},{"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":[[["1"],["3"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":" T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c: $ C = \\dfrac{1}{3-|x-2|} \\text { (bi\u1ebft C > 0) } $ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> min C = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"\u00c1p d\u1ee5ng: $|x| < a \\Rightarrow -a < x < a $","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> T\u00ccM GI\u00c1 TR\u1eca NH\u1ece NH\u1ea4T C\u1ee6A BI\u1ec2U TH\u1ee8C CH\u1ee8A D\u1ea4U GI\u00c1 TR\u1eca TUY\u1ec6T \u0110\u1ed0I <\/b> <br\/> $\\quad T = a + |f(x)| $ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1, $f(x)$ l\u00e0 bi\u1ec3u th\u1ee9c c\u00f3 ch\u1ee9a bi\u1ebfn $x$ <br\/> V\u00ec $|f(x)| \\geq 0 \\,\\,\\forall x \\in \\mathbb{Q}$ n\u00ean $T \\geq a$ <br\/> Khi \u0111\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $T$ b\u1eb1ng $a$ khi $f(x) = 0$ (ta ph\u1ea3i t\u00ecm $x$ \u0111\u1ec3 $f(x) = 0 $). <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <br\/> $ C = \\dfrac{1}{3-|x-2|}$ (bi\u1ebft C > 0) <br\/> C\u00f3 $C > 0 \\Rightarrow 3 - |x - 2| > 0 $ <br\/> $\\Rightarrow |x - 2| < 3 $ <br\/> $ \\Rightarrow -3 < x - 2 < 3 $ <br\/> $\\Rightarrow -1 < x < 5$ (*) <br\/> V\u00ec C > 0 n\u00ean C nh\u1ecf nh\u1ea5t khi $3 - |x -2|$ l\u00e0 s\u1ed1 d\u01b0\u01a1ng l\u1edbn nh\u1ea5t. <br\/> V\u00ec $|x - 2| \\geq 0 \\quad \\forall x \\in \\mathbb{Q}$ n\u00ean $ -|x - 2| \\leq 0 $ <br\/> $ \\Rightarrow 0 < 3 - |x - 2| \\leq 3 \\forall x \\in \\mathbb{Q} $ <br\/> $\\Rightarrow \\dfrac{1}{3-|x-2|} \\geq \\dfrac{1}{3},$ $\\forall x \\in \\mathbb{Q}$ <br\/> Hay $C \\geq \\dfrac{1}{3,}$ $\\forall x \\in \\mathbb{Q}$ <br\/> V\u1eady GTNN c\u1ee7a $C = \\dfrac{1}{3}$ khi $x - 2 = 0$ hay $x = 2$ (th\u1ecfa m\u00e3n (*)) <br\/> <br\/> <span class='basic_pink'> V\u1eady GTNN c\u1ee7a C l\u00e0 $\\dfrac{1}{3}$ khi $x = 2 $<\/span><\/span> <br\/>"}]}],"id_ques":192},{"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":[[["3"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":" T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c: $ B = |x + 1| + 2|6,9 - 3y| + 3 $ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> min B = _input_ ","hint":"\u00c1p d\u1ee5ng: $|f(x)| \\geq 0 \\forall x \\in \\mathbb{Q}$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> T\u00ccM GI\u00c1 TR\u1eca NH\u1ece NH\u1ea4T C\u1ee6A BI\u1ec2U TH\u1ee8C CH\u1ee8A D\u1ea4U GI\u00c1 TR\u1eca TUY\u1ec6T \u0110\u1ed0I <\/b> <br\/> $\\quad T = a + |f(x)| $ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1, $f(x)$ l\u00e0 bi\u1ec3u th\u1ee9c c\u00f3 ch\u1ee9a bi\u1ebfn $x$ <br\/> V\u00ec $|f(x)| \\geq 0\\,\\, \\forall x \\in \\mathbb{Q}$ n\u00ean $T \\geq a$ <br\/> Khi \u0111\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $T$ b\u1eb1ng $a$ khi $f(x) = 0$ (ta ph\u1ea3i t\u00ecm $x$ \u0111\u1ec3 $f(x) = 0 $). <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <br\/> $B = |x + 1| + 2|6,9 - 3y| + 3 $ <br\/> Ta c\u00f3: $|x + 1| \\geq 0,$ v\u1edbi $\\forall x \\in \\mathbb{Q}$ <br\/> $|6,9 - 3y| \\geq 0,$ v\u1edbi $\\forall y \\in \\mathbb{Q}$ <br\/> $\\Rightarrow B = |x + 1| + 2|6,9 - 3y| + 3 \\geq 0 + 2.0 + 3 = 3$ <br\/> Do \u0111\u00f3 GTNN c\u1ee7a $B = 3$ <br\/> D\u1ea5u ''='' x\u1ea3y ra $\\Leftrightarrow \\begin{cases} x + 1 = 0 \\Rightarrow x = -1 \\\\ 6,9 - 3y = 0 \\Rightarrow y = 6,9 : 3 = 2,3 \\end{cases} $<br\/> <br\/> <span class='basic_pink'> V\u1eady GTNN c\u1ee7a B l\u00e0 $3$ khi $x = -1; y = 2,3$<\/span><\/span> <br\/>"}]}],"id_ques":193},{"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":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai4/lv3/img\/2.jpg' \/><\/center> T\u00ecm $x$, bi\u1ebft: $ |x^{2} - 2x| = x $ ","select":["A. $ x = 0 $","B. $x \\in \\lbrace 0; 1; 3 \\rbrace$","C. $x \\in \\lbrace 1; 3 \\rbrace$ ","D. $x = 1$ "],"explain":"<span class='basic_left'> $ |x^{2} - 2x| = x $ <br\/> Ta c\u00f3 :$|x^{2} - 2x| \\geq 0$ n\u00ean $ x \\geq 0$ <br\/> Suy ra $|x^{2} - 2x| = |x(x - 2)| = x|x - 2|$ do $x \\geq 0$ <br\/> V\u1eady $ x|x - 2| = x \\Leftrightarrow x|x - 2| - x = 0 \\Leftrightarrow x(|x - 2| - 1) = 0 $ <br\/> $\\quad$ Th1 : $x = 0$ <br> $\\quad$ Th2: $|x - 2| - 1 = 0 \\Rightarrow |x - 2| = 1 \\\\ \\Rightarrow \\left[ \\begin{array}{} x - 2 = 1 \\Rightarrow x = 3 > 0 \\\\ x - 2 = -1 \\Rightarrow x = 1 > 0 \\end{array} \\right. $ <br\/> V\u1eady $x \\in \\lbrace 0; 1; 3 \\rbrace$ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> <br\/>","column":2}]}],"id_ques":194},{"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":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai4/lv3/img\/2.jpg' \/><\/center> T\u00ecm $x$, bi\u1ebft: $|x - 1| + |x - 4| = 3x $ ","select":["A. $ x = -5 $","B. $ x = -1 $","C. $x = 1$ ","D. A v\u00e0 C "],"hint":" X\u00e9t t\u1eebng kho\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a x \u0111\u1ec3 t\u00ecm d\u1ea5u gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i. ","explain":"<span class='basic_left'> Ta c\u00f3: $|x - 1| + |x - 4| = 3x $ <br\/> Ta x\u00e9t ba tr\u01b0\u1eddng h\u1ee3p: <br\/> * N\u1ebfu $ 0 \\leq x < 1$ th\u00ec $|x - 1| = -(x - 1) = 1 - x$ v\u00e0 $|x - 4| =-(x - 4) = 4 - x $ <br\/> Suy ra : $(1 - x) + (4 - x) = 3x \\Rightarrow 5 = 5x \\Rightarrow x = 1 > 0$ (lo\u1ea1i) <br\/> * N\u1ebfu $1 \\leq x \\leq 4$ th\u00ec $|x - 1| = x - 1$ v\u00e0 $|x - 4| =-(x - 4) = 4 - x $ <br\/> Suy ra : $(x - 1) + (4 - x) = 3x \\Rightarrow 3 = 3x \\Rightarrow x = 1$ (ch\u1ecdn) <br\/> * N\u1ebfu $x > 4$ th\u00ec $|x - 1| = x - 1$ v\u00e0 $|x - 4| = x - 4$ <br\/> Suy ra : $(x - 1) + (x - 4) = 3x \\Rightarrow -5 = x \\Rightarrow x = -5 < 4 $ (lo\u1ea1i) <br\/> V\u1eady $ x = 1$ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> <br\/>","column":2}]}],"id_ques":195},{"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":" T\u00ecm $x \\in \\mathbb{Z}$ v\u00e0 $x \\leq 2$ \u0111\u1ec3 bi\u1ec3u th\u1ee9c $N = x + \\dfrac{1}{2} - \\left| x - \\dfrac{2}{3} \\right| $ \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t ","select":["A. $ x = 3$","B. $ x = 1; x = 2 $","C. $ x = 2; x = 3 $","D. $ x = 4 $ "],"hint":" X\u00e9t hai tr\u01b0\u1eddng h\u1ee3p: $x - \\dfrac{2}{3} \\geq 0 $ v\u00e0 $ x - \\dfrac{2}{3} < 0 $ ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> T\u00ccM GI\u00c1 TR\u1eca L\u1edaN NH\u1ea4T C\u1ee6A BI\u1ec2U TH\u1ee8C CH\u1ee8A D\u1ea4U GI\u00c1 TR\u1eca TUY\u1ec6T \u0110\u1ed0I <\/b> <br\/> $\\quad T = b - |f(x)| $ v\u1edbi $b$ l\u00e0 h\u1eb1ng s\u1ed1, $f(x)$ l\u00e0 bi\u1ec3u th\u1ee9c c\u00f3 ch\u1ee9a bi\u1ebfn $x$ <br\/> V\u00ec $-|f(x)| \\leq 0 \\,\\, \\forall x \\in \\mathbb{Q}$ n\u00ean $T \\leq b$ <br\/> Khi \u0111\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a $T$ b\u1eb1ng $b$ khi $f(x) = 0.$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <br\/> Ta c\u00f3: $N = x + \\dfrac{1}{2} - \\left| x - \\dfrac{2}{3} \\right| $ <br\/> Ta x\u00e9t hai tr\u01b0\u1eddng h\u1ee3p: <br\/> *Tr\u01b0\u1eddng h\u1ee3p 1: N\u1ebfu $x - \\dfrac{2}{3} \\geq 0 \\Leftrightarrow x \\geq \\dfrac{2}{3}$ th\u00ec <br\/> $N = x + \\dfrac{1}{2} - (x - \\dfrac{2}{3}) = x + \\dfrac{1}{2} - x + \\dfrac{2}{3} = \\dfrac{7}{6} $(1) <br\/> *Tr\u01b0\u1eddng h\u1ee3p 2: N\u1ebfu $x < \\dfrac{2}{3}$ th\u00ec $N = x + \\dfrac{1}{2} + x - \\dfrac{2}{3} = \\dfrac{-1}{6} + 2x$ <br\/> V\u00ec $x < \\dfrac{2}{3}$ n\u00ean $2x < \\dfrac{4}{3}$ n\u00ean $N < \\dfrac{7}{6} $(2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $N \\leq \\dfrac{7}{6}$ <br\/> V\u1eady GTLN c\u1ee7a $N $ b\u1eb1ng $ \\dfrac{7}{6} \\Leftrightarrow x \\geq \\dfrac{2}{3}$ <br\/> M\u00e0 $x$ nguy\u00ean v\u00e0 $x \\leq 2$ n\u00ean $x = 1; x = 2 $ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> <br\/>","column":2}]}],"id_ques":196},{"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":[[["1"],["4"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":" T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c : $M = \\left| x + \\dfrac{1}{2} \\right| + \\left| x + \\dfrac{1}{3} \\right| + \\left| x + \\dfrac{1}{4} \\right| $ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> min M = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> T\u00ccM GI\u00c1 TR\u1eca NH\u1ece NH\u1ea4T C\u1ee6A BI\u1ec2U TH\u1ee8C CH\u1ee8A D\u1ea4U GI\u00c1 TR\u1eca TUY\u1ec6T \u0110\u1ed0I <\/b> <br\/> a) $\\quad T = a + |f(x)| $ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1, $f(x)$ l\u00e0 bi\u1ec3u th\u1ee9c c\u00f3 ch\u1ee9a bi\u1ebfn $x$ <br\/> V\u00ec $|f(x)| \\geq 0\\,\\, \\forall x \\in \\mathbb{Q}$ n\u00ean $T \\geq a$ <br\/> Khi \u0111\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $T$ b\u1eb1ng $a$ khi $f(x) = 0$ (ta ph\u1ea3i t\u00ecm $x$ \u0111\u1ec3 $f(x) = 0 $). <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <br\/> $M = \\left| x + \\dfrac{1}{2} \\right| + \\left| x + \\dfrac{1}{3} \\right| + \\left| x + \\dfrac{1}{4} \\right| $ <br\/> Ta c\u00f3: $ \\left| x + \\dfrac{1}{4} \\right| = \\left| -x - \\dfrac{1}{4} \\right| \\geq -x - \\dfrac{1}{4} $ <br\/> $ \\left| x + \\dfrac{1}{3} \\right| \\geq 0$ <br\/> $ \\left| x + \\dfrac{1}{2} \\right| \\geq x + \\dfrac{1}{2}$ <br\/> Do \u0111\u00f3: $M \\geq x + \\dfrac{1}{2} + 0 - x - \\dfrac{1}{4} = \\dfrac{1}{4}$ <br\/> D\u1ea5u ''='' x\u1ea3y ra $\\Leftrightarrow x + \\dfrac{1}{4} \\leq 0 ; x + \\dfrac{1}{3} = 0 ; x + \\dfrac{1}{2} \\geq 0$ <br\/> $\\Leftrightarrow x = -\\dfrac{1}{3}$<br\/> <br\/> <span class='basic_pink'> V\u1eady GTNN c\u1ee7a M l\u00e0 $\\dfrac{1}{4}$ khi $x = \\dfrac{-1}{3}$<\/span><\/span> <br\/>"}]}],"id_ques":197},{"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":" T\u00ednh $ H = 1,25 . |1,4x - 1,01| - 3,2(x : 0,25 - 12,15) $ v\u1edbi $x = 2,35$ ","select":["A. $ 11,65$","B. $ 5,95 $","C. $ -5,65 $","D. $ 74,335 $ "],"hint":" Thay $x = 2,35$ v\u00e0o H r\u1ed3i th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh. ","explain":" Thay $x = 2,35$ v\u00e0o H ta \u0111\u01b0\u1ee3c: <br\/> $ H = 1,25 . |1,4 . 2,35 - 1,01| - 3,2 . (2,35 : 0,25 - 12,15) \\\\ = 1,25 . |3,29 - 1,01| - 3,2 . (9,4 - 12,15) \\\\ = 1,25 . |2,28| + 3,2 . (12,15 - 9,4) \\\\ = 1,25 . 2,28 + 3,2 . 2,75 \\\\ = 2,85 + 8,8 = 11,65$ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span><\/span> <br\/>","column":2}]}],"id_ques":198},{"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":" T\u00ecm $x$ v\u00e0 $y$, bi\u1ebft: $|x - y| + |y + 0,28| = 0 $ ","select":["A. $ x = 0,28 $","B. $ x = -0,28 $","C. $ x = y = 0,28 $","D. $ x = y = -0,28 $ "],"hint":" $|x - y| + |y + 0,28| = 0 \\Leftrightarrow |x - y| = |y + 0,28| = 0$ ","explain":" $|x - y| + |y + 0,28| = 0 $ <br\/> Ta c\u00f3 :$|x - y| \\geq 0$ v\u00e0 $|y + 0,28| \\geq 0 $ <br\/> M\u00e0 theo b\u00e0i $ |x - y| + |y + 0,28| = 0 $, n\u00ean $|x - y| = |y + 0,28| = 0$ <br\/> Suy ra: $\\begin{cases} x - y = 0 \\\\ y + 0,28 = 0 \\end{cases} \\Rightarrow \\begin{cases} x = -0,28 \\\\ y = -0,28 \\end{cases} $ <br\/> V\u1eady $ x = y = -0,28 $<br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> <br\/>","column":2}]}],"id_ques":199},{"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":[[["4"],["11"],["-5"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":" T\u00ecm $x$ v\u00e0 $y$ bi\u1ebft r\u1eb1ng: <br\/> $ | x - \\dfrac{4}{11}| + |5 + y| = 0 ; x = $ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>$ y = _input_ ","hint":"$ \\left| x - \\dfrac{4}{11}\\right| + |5 + y| = 0 \\Leftrightarrow \\left| x - \\dfrac{4}{11}\\right| = |5 + y| = 0$","explain":" $ \\left| x - \\dfrac{4}{11}\\right| + |5 + y| = 0$ <br\/> Ta c\u00f3: $\\left| x - \\dfrac{4}{11}\\right| \\geq 0; |5 + y| \\geq 0 \\quad \\forall x,y \\in \\mathbb{Q}$ <br\/> m\u00e0 $\\left| x - \\dfrac{4}{11}\\right| + |5 + y| = 0$ <br\/> n\u00ean $\\left| x - \\dfrac{4}{11}\\right| = 0$ v\u00e0 $|5 + y| = 0 $ <br\/> Suy ra: $\\begin{cases} x - \\dfrac{4}{11} = 0 \\Rightarrow x = \\dfrac{4}{11} \\\\ 5 + y = 0 \\Rightarrow y = -5 \\end{cases} $ <br\/> <br\/> <span class='basic_pink'> V\u1eady $x = \\dfrac{4}{11}; \\quad y = -5 $.<\/span><\/span> <br\/>"}]}],"id_ques":200}],"lesson":{"save":0,"level":3}}