Bài 1:
a: ĐKXĐ: x-2<>0
=>x<>2
b: ĐKXĐ: \(x^2-6x<>0\)
=>x(x-6)<>0
=>x∉{0;6}
c: ĐKXĐ: \(3x^2-4x<>0\)
=>x(3x-4)<>0
=>x∉{0;4/3}
Bài 2:
a: \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+1\right)+2\left(2x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{x^2+5x+6}{2x\left(x+3\right)}=\frac{\left(x+3\right)\left(x+2\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x-x+6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}\)
\(=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c: \(\frac{2x+6}{3x^2-x}:\frac{x^2+3x}{1-3x}\)
\(=\frac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\frac{-\left(3x-1\right)}{x\left(x+3\right)}=\frac{-2}{x^2}\)
