Lời giải:

+) Với \(x>0\) thì \(\text{BPT}\Leftrightarrow \frac{2}{x}\geq 2\Leftrightarrow x\leq 1\)

Nghiệm: \(x\in (0,1]\)

+) Với \(0>x\geq -2\)

\(\text{BPT}\Leftrightarrow \frac{2}{x}\geq 2\Leftrightarrow x\geq 1(\text{vô lý})\)

+) Với \(x<-2\)

\(\text{BPT}\Leftrightarrow \frac{-2x-2}{x}\geq 2\Leftrightarrow -2x-2\leq 2x\Leftrightarrow x\geq \frac{-1}{2}\)( vô lý)

Lưu ý: Khi $x<0$ thì khi nhân lên triệt tiêu mẫu phải đổi dấu

Vậy nghiệm của BPT là \(x\in (0,1]\)

   \(\frac{2x-2}{3}\ge2-\frac{x+2}{2}\)

\(\Leftrightarrow\frac{2x-2}{3}\ge\frac{6-x}{2}\)

\(\Rightarrow2.\left(2x-2\right)=3.\left(6-x\right)\)

\(\Leftrightarrow4x-4=18-3x\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\frac{22}{7}\)

Học Tốt Nha!!

\(\frac{2x-2}{3}\ge2-\frac{x+2}{2}\)

\(\Leftrightarrow\frac{2\left(2x-2\right)}{6}\ge\frac{12-3\left(x+2\right)}{6}\)

\(\Leftrightarrow4x-4\ge12-3x-6\)

\(\Leftrightarrow4x+3x\ge6+4\)

\(\Leftrightarrow7x\ge10\)

\(\Leftrightarrow x\ge\frac{10}{7}\)

ĐKXĐ: \(x\le2;x\ne0\)

- Với \(0< x\le2\)

\(\Leftrightarrow4x-3+\sqrt{2-x}\ge2x\)

\(\Leftrightarrow\sqrt{2-x}\ge3-2x\)

+ Với \(x>\frac{3}{2}\) BPT luôn đúng

+ Với \(x\le\frac{3}{2}\Leftrightarrow2-x\ge4x^2-12x+9\)

\(\Leftrightarrow4x^2-11x+7\le0\Rightarrow1\le x\le\frac{7}{4}\) \(\Rightarrow1\le x\le\frac{3}{2}\)

- Với \(x< 0\Leftrightarrow4x-3+\sqrt{2-x}\le2x\)

\(\Leftrightarrow3-2x\ge\sqrt{2-x}\)

\(\Leftrightarrow\left(3-2x\right)^2\ge2-x\Leftrightarrow4x^2-11x+7\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow x< 0\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}x< 0\\1\le x\le2\end{matrix}\right.\)

Dk 1<x<2

√x^2 -x -2<x+2

5x+6>0

X > -6/5

Bpt vô nghiệm

Đk: \(x\ge\dfrac{1}{2}\)

Bpt\(\Leftrightarrow\left(x^2+2x\sqrt{2x-1}+2x-1\right)-\left[4\left(2x-1\right)+4\sqrt{2x-1}+1\right]\ge0\)

\(\Leftrightarrow\left(x+\sqrt{2x-1}\right)^2-\left(2\sqrt{2x-1}+1\right)^2\ge0\)

\(\Leftrightarrow\left(x-\sqrt{2x-1}-1\right)\left(x+3\sqrt{2x-1}+1\right)\ge0\) (1)

Vì \(x\ge\dfrac{1}{2}\Rightarrow x+3\sqrt{2x-1}+1>0\)

Từ (1) \(\Rightarrow x-\sqrt{2x-1}-1\ge0\)

\(\Leftrightarrow\sqrt{2x-1}\le x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\x-1\ge0\\2x-1\le\left(1-x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\in R\backslash\left(2-\sqrt{2};2+\sqrt{2}\right)\end{matrix}\right.\)\(\Rightarrow x\ge2+\sqrt{2}\)

Vậy...

ĐKXĐ: \(\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)

\(\frac{\left|2x-1\right|}{\left(x+1\right)\left(x-2\right)}>\frac{1}{2}\) (*)

+) Nếu \(x>2\) thì (*) \(\Leftrightarrow\frac{2x-1}{x^2-x-2}>\frac{1}{2}\)

\(\Leftrightarrow4x-2>x^2-x-2\)

\(\Leftrightarrow x^2-5x< 0\)

\(\Leftrightarrow x\left(x-5\right)< 0\)

\(\Leftrightarrow0< x< 5\)

\(\Leftrightarrow2< x< 5\)

+) Nếu \(x< -1\) thì (*) \(\Leftrightarrow\frac{1-2x}{x^2-x-2}>\frac{1}{2}\)

\(\Leftrightarrow2-4x>x^2-x-2\)

\(\Leftrightarrow x^2+3x-4< 0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)< 0\)

\(\Leftrightarrow-4< x< 1\)

\(\Leftrightarrow-4< x< -1\)

Vậy...

\(\frac{x}{x-2}+\frac{x+2}{x}>2\)

\(\frac{x^2}{x\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}>2\)

\(\frac{x^2+x^2-4}{x\left(x-2\right)}>2\)

\(2x^2-4>2.x\left(x-2\right)\)

\(x^2-2>x^2-2x\)

\(\Leftrightarrow2>2x\)

\(\Rightarrow x< 1\)

\(\frac{x^2}{x\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}>\frac{2x\left(x-2\right)}{x\left(x-2\right)}\)

=\(\frac{x^2+x^2-4}{x\left(x-2\right)}>\frac{2x^2+2x}{x\left(x-2\right)}\)

=>\(x^2+x^2-4>2x^2+2x\)

= \(x^2+x^2-2x^2-2x>4\)

=-2x>4

=x<-2

thick cái nha