Bài 2:
a: ĐKXĐ: x∉{0;2}
Ta có: \(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)
=>\(\frac{x+2}{x-2}-\frac{2}{x\left(x-2\right)}=\frac{1}{x}\)
=>\(\frac{x\left(x+2\right)-2}{x\left(x-2\right)}=\frac{x-2}{x\left(x-2\right)}\)
=>\(x^2+2x-2=x-2\)
=>\(x^2+x=0\)
=>x(x+1)=0
=>\(\left[\begin{array}{l}x=0\left(loại\right)\\ x=-1\left(nhận\right)\end{array}\right.\)
b: \(\frac{x-2}{3}-x\ge\frac{2x+1}{2}+1\)
=>\(\frac{x-2-3x}{3}\ge\frac{2x+1+2}{2}\)
=>\(\frac{-2x-2}{3}\ge\frac{2x+3}{2}\)
=>2(-2x-2)>=3(2x+3)
=>-4x-4>=6x+9
=>-10x>=13
=>\(x\le-\frac{13}{10}\)
1:
a: \(\frac23\cdot\sqrt9-\frac32\cdot\sqrt{\left(-6\right)^2}+7\)
\(=\frac23\cdot3-\frac32\cdot6+7\)
=2-9+7
=0
b: \(\sqrt{\left(5+\sqrt7\right)^2}-\sqrt{8-2\sqrt7}\)
\(=5+\sqrt7-\sqrt{\left(\sqrt7-1\right)^2}\)
\(=5+\sqrt7-\left(\sqrt7-1\right)=5+\sqrt7-\sqrt7+1=6\)
