bn dùng pp chia khoảng cho nhanh nhé
c.
\(\left(3x^2-4x\right)\left(2x^2-x-1\right)=0\Rightarrow x=\dfrac{4}{3};-\dfrac{1}{2};0;1\)
Vậy x < -1/2 ; 0 < x < 1
d,
\(\left(4x^2-1\right)\left(-8x^2+x-3\right)\left(2x+9\right)=0\Leftrightarrow x=-\dfrac{1}{2};\dfrac{1}{2};-\dfrac{9}{2}\)
Vậy x < -9/2 ; -1/2 < x < 1/2
e, \(\dfrac{4x^2+3x-1}{x^2+5x+7}\ge0\)
Ta có \(x^2+5x+7=x^2+\dfrac{2.5}{2}x+\dfrac{25}{4}+\dfrac{3}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Rightarrow4x^2+3x-1\ge0\Leftrightarrow\left(4x-1\right)\left(x+1\right)\ge0\)
TH1 : \(\left\{{}\begin{matrix}4x-1\ge0\\x+1\ge0\end{matrix}\right.\Leftrightarrow x\ge\dfrac{1}{4}\)
TH2 : \(\left\{{}\begin{matrix}4x-1\le0\\x+1\le0\end{matrix}\right.\Leftrightarrow x\le-1\)
f, \(\dfrac{5x^2-7x-3}{3x^2-2x-5}-1>0\Leftrightarrow\dfrac{5x^2-7x-3-3x^2+2x+5}{3x^2-2x-5}>0\)
\(\Leftrightarrow\dfrac{2x^2-5x+2}{3x^2-2x-5}>0\Leftrightarrow\dfrac{\left(x-2\right)\left(2x-1\right)}{\left(3x-5\right)\left(x+1\right)}>0\)
-bn tự xét tưng th nhé
giúp mình câu b) c) với ạ
B. 
C. 
D. 
