1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~

\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)

Mấy bài kia sao cái phương trình dài thê,s giải sao nổi

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

\(ĐK:\hept{\begin{cases}x-1\ne0\\x+3\ne\\x^2+2x-3\ne0\end{cases}0}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\Leftrightarrow-3\end{cases}}\)

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)+4-x^2-2x+3=0\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5+4-x^2-2x+3=0\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\Leftrightarrow x=-3\) (loại)

 Vậy pt vô N^o

Số t tính đc rất thú dị :) 

a/ \(2x^2-3x+1>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{2}\end{matrix}\right.\)

b/ \(-3x^2+2x+1< 0\Rightarrow-\frac{1}{3}< x< 1\)

c/ \(\frac{x+3}{x-2}\ge0\Rightarrow\left[{}\begin{matrix}x>2\\x\le-3\end{matrix}\right.\)

d/ \(\frac{2x+1}{x+2}\ge1\Leftrightarrow\frac{2x+1}{x+2}-1\ge0\Leftrightarrow\frac{x-1}{x+2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge1\\x< -2\end{matrix}\right.\)

e/ \(\frac{\sqrt{x}+3}{2-\sqrt{x}}\le0\Rightarrow\left\{{}\begin{matrix}x\ge0\\2-\sqrt{x}< 0\end{matrix}\right.\) \(\Rightarrow x>4\)

g/\(\frac{\sqrt{x}-3}{\sqrt{x}-2}\ge0\Rightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x\ge9\\x< 4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.\)

h/ \(\frac{\sqrt{x}-3}{\sqrt{x}-1}-\frac{1}{3}< 0\Rightarrow\frac{2\left(\sqrt{x}-4\right)}{3\left(\sqrt{x}-1\right)}< 0\Rightarrow1< x< 16\)

ĐKXĐ: x∉{1;2}

Ta có: \(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\)

\(\Leftrightarrow\frac{\left(2x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\frac{\left(3x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4\left(x^2-3x+2\right)}{\left(x-1\right)\left(x-2\right)}\)

Suy ra: \(\left(2x-1\right)\left(x-2\right)+\left(3x-2\right)\left(x-1\right)=x^2+4x+5+4\left(x^2-3x+2\right)\)

\(\Leftrightarrow2x^2-4x-x+2+3x^2-3x-2x+2=x^2+4x+5+4x^2-12x+8\)

\(\Leftrightarrow5x^2-10x+4=5x^2-8x+13\)

\(\Leftrightarrow5x^2-10x+4-5x^2+8x-13=0\)

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow-2x=9\)

hay \(x=\frac{-9}{2}\)(tm)

Vậy: \(S=\left\{-\frac{9}{2}\right\}\)

ĐKXĐ: \(x\ne1;x\ne2\)

\(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\\ \Leftrightarrow\frac{2x^2-5x+2}{\left(x-1\right)\left(x-2\right)}+\frac{3x^2-5x+2}{\left(x-1\right)\left(x-2\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4x^2-12x+8}{\left(x-1\right)\left(x-2\right)}\\ \Rightarrow2x^2-5x+2+3x^2-5x+2=x^2+4x+5+4x^2-12x+8\\ \Leftrightarrow2x^2+3x^2-x^2-4x^2-5x-5x-4x+12x=5-2-2\\ \Leftrightarrow-2x=1\\ \Leftrightarrow x=\frac{-1}{2}\left(tm\right)\)Vậy tập nghiệm của phương trình là: \(S=\left\{-\frac{1}{2}\right\}\)