{"segment":[{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai25/lv3/img\/5.jpg' \/><\/center><br\/>So s\u00e1nh $a$ v\u00e0 $b$ n\u1ebfu: $-15a + 12 \\geq -15b + 12$","select":[" A. $a \\leq b $"," B. $a \\geq b$","C. $a > b$","D. $a < b$"],"hint":"S\u1eed d\u1ee5ng quy t\u1eafc c\u1ed9ng v\u00e0 nh\u00e2n c\u1ea3 hai v\u1ebf b\u1ea5t \u0111\u1eb3ng th\u1ee9c cho c\u00f9ng m\u1ed9t s\u1ed1.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>+ Khi c\u1ed9ng c\u00f9ng m\u1ed9t s\u1ed1 v\u00e0o c\u1ea3 hai v\u1ebf c\u1ee7a m\u1ed9t b\u1ea5t \u0111\u1eb3ng th\u1ee9c ta \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c m\u1edbi c\u00f9ng chi\u1ec1u v\u1edbi b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho<br\/>+ Khi nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a m\u1ed9t b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 d\u01b0\u01a1ng ta \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c m\u1edbi c\u00f9ng chi\u1ec1u v\u1edbi b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho<\/span> <br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $-15a + 12 \\geq -15b + 12$<br\/>$\\Rightarrow -15a + 12 + (-12) \\geq -15b + 12 + (-12)$<br\/>$\\Rightarrow -15a \\geq -15b \\Rightarrow -15a.\\dfrac{1}{-15} \\leq -15b.\\dfrac{1}{-15} \\Rightarrow a \\leq b$ <br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>A. $a \\leq b$<\/span> ","column":2}]}],"id_ques":1091},{"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 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c: $A = x^6 + y^6$, bi\u1ebft $x^2 + y^2 = 1$.","select":[" A. max$A = 0$"," B. max$A = 1$","C. max$A = -1$","D. max$A = -3$"],"hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c t\u1ed5ng c\u1ee7a hai l\u1eadp ph\u01b0\u01a1ng \u0111\u1ec3 khai tri\u1ec3n $A$.","explain":"<span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align*}A& = x^6 + y^6 \\\\&= (x^2)^3 + (y^2)^3 \\\\&= (x^2 + y^2)(x^4 -x^2y^2 + y^4) \\\\&= x^4 - x^2y^2 + y^4 \\\\&= (x^4 + 2x^2y^2 + y^4) - 3x^2y^2 \\\\&= (x^2 + y^2)^2 - 3x^2y^2 \\\\&= 1 - 3x^2y^2\\end{align*}$<br\/>L\u1ea1i c\u00f3: $-3x^2y^2 \\leq 0 \\ \\forall x \\Rightarrow 1 - 3x^2y^2 \\leq 1 \\forall x$<br\/>V\u1eady gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a $A$ l\u00e0 $1$<br\/>D\u1ea5u \"=\" x\u1ea3y ra khi $x = 0$ ho\u1eb7c $y = 0$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>B. max$A = 1$<\/span>","column":2}]}],"id_ques":1092},{"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 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c<br\/>$A = x^2 - x$","select":[" A. min$A = \\dfrac{1}{2}$"," B. min$A = \\dfrac{-1}{2}$","C. min$A = \\dfrac{1}{4}$","D. min$A = \\dfrac{-1}{4}$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng $[f(x)]^2 + a$<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A$<\/span> <br\/><br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align*}A& = x^2 -x \\\\&= x^2 -2x\\dfrac{1}{2} + \\left(\\dfrac{1}{2}\\right)^2 - \\left(\\dfrac{1}{2}\\right)^2 \\\\&= \\left(x - \\dfrac{1}{2}\\right)^2 - \\dfrac{1}{4}\\end{align*}$<br\/>L\u1ea1i c\u00f3: $\\left(x - \\dfrac{1}{2}\\right)^2 \\geq 0 \\ \\forall x \\Rightarrow \\left(x - \\dfrac{1}{2}\\right)^2 - \\dfrac{1}{4} \\geq -\\dfrac{1}{4} \\forall x$<br\/>V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A$ l\u00e0 $-\\dfrac{1}{4}$<br\/>D\u1ea5u \"=\" x\u1ea3y ra khi $x - \\dfrac{1}{2} = 0 \\Leftrightarrow x = \\dfrac{1}{2}$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>D. min$A = -\\dfrac{1}{4}$<\/span>","column":2}]}],"id_ques":1093},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>,<) v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["<"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>So s\u00e1nh $\\dfrac{1}{1.3} + \\dfrac{1}{3.5} + ... + \\dfrac{1}{(2n - 1)(2n + 1)} $ v\u00e0 $\\dfrac{1}{2}$ <br\/><b>\u0110\u00e1p \u00e1n:<\/b> $\\dfrac{1}{1.3} + \\dfrac{1}{3.5} + ... + \\dfrac{1}{(2n - 1)(2n + 1)} $ _input_ $\\dfrac{1}{2}$<\/span>","hint":"$\\dfrac{m - n}{n.m} = \\dfrac{1}{n} - \\dfrac{1}{m}$.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align*}& \\dfrac{1}{1.3} + \\dfrac{1}{3.5} + ... + \\dfrac{1}{(2n - 1)(2n + 1)} \\\\&= \\dfrac{1}{2} \\left(\\dfrac{2}{1.3} + \\dfrac{2}{3.5} + ... + \\dfrac{2}{(2n - 1)(2n + 1)} \\right) \\\\&= \\dfrac{1}{2} \\left(1 - \\dfrac{1}{3} + \\dfrac{1}{3} - \\dfrac{1}{5 }+ ... + \\dfrac{1}{2n - 1} - \\dfrac{1}{2n + 1}\\right) \\\\&= \\dfrac{1}{2} \\left(1 - \\dfrac{1}{2n + 1}\\right) \\end{align*}$<br\/>L\u1ea1i c\u00f3: $1 - \\dfrac{1}{2n + 1} = \\dfrac{2n + 1}{2n + 1} - \\dfrac{1}{2n + 1}= \\dfrac{2n + 1 - 1}{2n + 1} = \\dfrac{2n}{2n + 1} < 1$<br\/>$\\Rightarrow \\dfrac{1}{2} \\left(1 - \\dfrac{1}{2n + 1}\\right) < \\dfrac{1}{2}$<br\/>Hay $\\dfrac{1}{1.3} + \\dfrac{1}{3.5} + ... + \\dfrac{1}{(2n - 1)(2n + 1)} < \\dfrac{1}{2}$<br\/>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>\"<\"<\/span><\/span>"}]}],"id_ques":1094},{"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 $0 < a < b$, h\u00e3y so s\u00e1nh: $a^3$ v\u00e0 $b^3$","select":[" A. $a^3 \\geq b^3$"," B. $a^3 = b^3$","C. $a^3 > b^3$","D. $a^3 < b^3$"],"hint":"S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t b\u1eafc c\u1ea7u: c\u00f9ng so s\u00e1nh v\u1edbi $a^2b$ v\u00e0 $ab^2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Khi nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 d\u01b0\u01a1ng ta \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c m\u1edbi c\u00f9ng chi\u1ec1u v\u1edbi b\u1eaft \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho<\/span> <br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>+) $a < b$ v\u00e0 $a^2 > 0$ $ \\Rightarrow a.a^2 < b.a^2 \\Rightarrow a^3 < a^2b$ <b>(1)<\/b><br\/>+) $a < b$ v\u00e0 $b^2 > 0$ $\\Rightarrow a.b^2 < b.b^2 \\Rightarrow ab^2 < b^3$ <b>(2)<\/b><br\/>+) $a < b$ v\u00e0 $a.b > 0 \\ \\text{v\u00ec} \\ 0 < a < b$ $\\Rightarrow a.ab < b.ab \\Rightarrow a^2b < ab^2$ <b>(3)<\/b><br\/>T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow a^3 < b^3$.<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'> D. $a^3 < b^3$<\/span><\/span> <br\/><\/span>","column":2}]}],"id_ques":1095},{"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":"Cho $11x - 2 > 11y + 3$, h\u00e3y so s\u00e1nh: $x$ v\u00e0 $y$","select":[" A. $x \\geq y$"," B. $x \\leq y$","C. $x > y$","D. $x < y$"],"hint":"S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t b\u1eafc c\u1ea7u: c\u00f9ng so s\u00e1nh v\u1edbi $11y - 2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Khi nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 d\u01b0\u01a1ng ta \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c m\u1edbi c\u00f9ng chi\u1ec1u v\u1edbi b\u1eaft \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho<br\/>+) Khi c\u1ed9ng c\u1ea3 hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 ta \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c m\u1edbi c\u00f9ng chi\u1ec1u v\u1edbi b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho<\/span> <br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: $3 > -2$ $ \\Rightarrow 11y + 3 > 11y - 2$ <b>(1)<\/b><br\/>L\u1ea1i c\u00f3: $11x - 2 > 11y + 3$ <b>(2)<\/b><br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow 11x - 2 > 11y - 2 \\\\ \\Rightarrow 11x -2 + 2 > 11y - 2 + 2 \\\\ \\Rightarrow 11x > 11y \\\\ \\Rightarrow 11x\\dfrac{1}{11} > 11y\\dfrac{1}{11} \\\\ \\Rightarrow x > y$.<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'> C. $x > y$<\/span><\/span> <br\/><\/span>","column":2}]}],"id_ques":1096},{"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":"Cho $a, b > 0$ th\u1ecfa m\u00e3n: $a + b = 1$. So s\u00e1nh $a^3 + b^3$ v\u00e0 $\\dfrac{1}{4}$","select":[" A. $a^3 + b^3 \\geq \\dfrac{1}{4}$"," B. $a^3 + b^3 \\leq \\dfrac{1}{4}$","C. $a^3 + b^3 > \\dfrac{1}{4}$","D. $a^3 + b^3 < \\dfrac{1}{4}$"],"hint":"Ph\u00e2n t\u00edch $a^3 + b^3$ v\u00e0 k\u1ebft h\u1ee3p s\u1eed d\u1ee5ng b\u1ea5t \u0111\u0103ng th\u1ee9c Cosi \u0111\u1ec3 gi\u1ea3i.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align*}a^3 + b^3 &= (a + b)(a^2 -ab + b^2) \\\\&= 1.(a^2 - ab + b^2) \\\\& = a^2 + 2ab + b^2 - 3ab \\\\&= (a + b)^2 - 3ab \\\\&= 1 - 3ab \\end{align*}$<br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cosi cho $a, b > 0$, ta c\u00f3: <br\/>$a + b \\geq 2 \\sqrt{ab}$<br\/>$\\Rightarrow ab \\leq \\left(\\dfrac{a + b}{2}\\right)^2$<br\/> $\\Rightarrow ab \\leq \\left(\\dfrac{1}{2}\\right)^2$ hay $ ab \\leq \\dfrac{1}{4}$<br\/>$\\Rightarrow -3.ab \\geq -3.\\dfrac{1}{4} \\Rightarrow 1-3ab \\geq 1-3.\\dfrac{1}{4} = \\dfrac{1}{4}$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0<span class='basic_pink'> A. $a^3 + b^3 \\geq \\dfrac{1}{4}$<\/span> <br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/> B\u1ea5t \u0111\u1eb3ng th\u1ee9c Cosi: $a, b > 0$ th\u00ec $a + b \\geq 2 \\sqrt{ab}$<br\/>Ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cosi:<br\/>V\u1edbi $a, b > 0$ th\u00ec:<br\/> $(\\sqrt{a} - \\sqrt{b})^2 \\geq 0$<br\/>$\\Rightarrow \\sqrt{a}^2 - 2\\sqrt{ab} + \\sqrt{b}^2 \\geq 0$<br\/> $\\Rightarrow a + b \\geq 2\\sqrt{ab}$ (\u0111pcm) <br\/><\/span> <br\/><br\/>","column":2}]}],"id_ques":1097},{"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":"Cho $a, b, c > 0$. So s\u00e1nh $\\dfrac{a}{b} + \\dfrac{b}{a}$ v\u00e0 $2$","select":[" A. $\\dfrac{a}{b} + \\dfrac{b}{a} \\geq 2$"," B. $\\dfrac{a}{b} + \\dfrac{b}{a} \\leq 2$","C. $\\dfrac{a}{b} + \\dfrac{b}{a} > 2$","D. $\\dfrac{a}{b} + \\dfrac{b}{a} < 2$"],"hint":"Ph\u00e2n t\u00edch $\\dfrac{a}{b} + \\dfrac{b}{a}$ k\u1ebft h\u1ee3p s\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cosi.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\dfrac{a}{b} + \\dfrac{b}{a} = \\dfrac{a^2 + b^2}{ab}$<br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cosi cho $a, b > 0$, ta c\u00f3: <br\/>$a^2 + b^2 \\geq 2ab$<br\/>$\\Rightarrow \\dfrac{a^2 + b^2}{ab} \\geq \\dfrac{2ab}{ab}$<br\/> $\\Rightarrow \\dfrac{a^2 + b^2}{ab} \\geq 2$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0<span class='basic_pink'> A. $\\dfrac{a}{b} + \\dfrac{b}{a} \\geq 2$<\/span> <br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/> B\u1ea5t \u0111\u1eb3ng th\u1ee9c Cosi: $a, b > 0$ th\u00ec $a + b \\geq 2 \\sqrt{ab}$<br\/>Ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cosi:<br\/>V\u1edbi $a, b > 0$ th\u00ec:<br\/> $(\\sqrt{a} - \\sqrt{b})^2 \\geq 0$<br\/>$\\Rightarrow \\sqrt{a}^2 - 2\\sqrt{ab} + \\sqrt{b}^2 \\geq 0$<br\/> $\\Rightarrow a + b \\geq 2\\sqrt{ab}$ (\u0111pcm) <br\/><\/span> <br\/><br\/>","column":2}]}],"id_ques":1098},{"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 $5x < 8x$. So s\u00e1nh $x$ v\u00e0 $0$","select":[" A. $x < 0$"," B. $x > 0$","C. x = 0"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $5x < 8x$ <br\/>L\u1ea1i c\u00f3: $5 < 8$ $\\Rightarrow x > 0$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0<span class='basic_pink'> A. $x < 0$<\/span> ","column":3}]}],"id_ques":1099},{"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":"S\u1ed1 $a$ l\u00e0 s\u1ed1 \u00e2m hay s\u1ed1 d\u01b0\u01a1ng n\u1ebfu: $-5a - 7 < -9a - 7$","select":[" A. S\u1ed1 d\u01b0\u01a1ng"," B. S\u1ed1 \u00e2m"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>+ Khi nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 d\u01b0\u01a1ng ta \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c m\u1edbi c\u00f9ng chi\u1ec1u v\u1edbi b\u1eaft \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho<br\/>+ Khi nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 \u00e2m ta \u0111\u01b0\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c m\u1edbi ng\u01b0\u1ee3c chi\u1ec1u v\u1edbi b\u1eaft \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho<\/span> <br\/><br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: $-5a - 7 < -9a - 7$<br\/>$\\Rightarrow -5a - 7 + 7 < -9a - 7 + 7 \\\\ \\Rightarrow -5a < -9a$<br\/>M\u00e0 $-5 > -9$<br\/>$ \\Rightarrow a < 0$ hay $a$ l\u00e0 s\u1ed1 \u00e2m<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>B. S\u1ed1 \u00e2m<\/span>","column":2}]}],"id_ques":1100}],"lesson":{"save":0,"level":3}}