Question

giải bpt

1.\(\frac{1}{x+2}\ge\frac{x+2}{3x-5}\)

2.\(\sqrt{-x^2+4x-3}\le x-1\)

help!

Answer 1

\(\frac{1}{x+2}-\frac{x+2}{3x-5}\ge0\)

\(\Leftrightarrow\frac{-x^2-x-9}{\left(x+2\right)\left(3x-5\right)}\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-5\right)< 0\) (do \(-x^2-x-9< 0;\forall x\))

\(\Rightarrow-2< x< \frac{5}{3}\)

2/ ĐKXĐ: \(1\le x\le3\)

\(\Leftrightarrow-x^2+4x-3\le\left(x-1\right)^2\)

\(\Leftrightarrow2x^2-6x+4\ge0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le1\end{matrix}\right.\)

Kết hợp ĐKXĐ: \(\left[{}\begin{matrix}x=1\\2\le x\le3\end{matrix}\right.\)