Question

[Nguyễn Khánh Linh]

Câu 4 : Giải các phương trình sau

a , \(\)\(\frac{-25}{\left(-x+2\right)\left(-3x-2\right)}\)< 0

b , \(\frac{1}{x-1}>\frac{2}{2x-1}\)

c , \(\frac{2}{-x+3}+\frac{1}{\frac{1}{2}x-\frac{3}{2}}\le0\)

d , \(\frac{x-1}{x^2-3x+2}\le1\)

e , \(\frac{x+1}{x^2+x+2}>\frac{1}{x+1}\)

Answer 1

[Nguyễn Việt Lâm]

a/ \(\frac{-25}{\left(-x+2\right)\left(-3x-2\right)}< 0\Leftrightarrow\left[{}\begin{matrix}x< -\frac{2}{3}\\x>2\end{matrix}\right.\)

b/ \(\frac{1}{x-1}-\frac{2}{2x-1}>0\Leftrightarrow\frac{1}{\left(x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{2}\end{matrix}\right.\)

c/ \(\frac{2}{3-x}+\frac{2}{x-3}\le0\Leftrightarrow0\le0\) (luôn đúng)

Vậy nghiệm của BPT là \(R\backslash\left\{3\right\}\)

d/ \(1-\frac{x-1}{x^2-3x+2}\ge\Leftrightarrow\frac{x^2-4x+3}{\left(x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\ge3\\1< x< 2\\x< 1\end{matrix}\right.\)

e/ \(\frac{x+1}{x^2+x+2}-\frac{1}{x+1}>0\Leftrightarrow\frac{x-1}{\left(x+1\right)\left(x^2+x+2\right)}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)