a) Ta có: \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+1\right)}{2x\left(x+3\right)}+\frac{2\cdot\left(2x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+5x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}\)
\(=\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}\)
\(=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) Ta có: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(=\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x}{x\left(2x+6\right)}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x-x+6}{x\left(2x+6\right)}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)
c) Ta có: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)\cdot2\cdot\left(2-x\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}\)
\(=\frac{5\cdot2\cdot\left(2-x\right)}{-4\left(2-x\right)}=\frac{5\cdot2}{-4}=\frac{-5}{2}\)
d) Ta có: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3x}{x\left(x+4\right)\cdot2\left(2-x\right)}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3}{2\left(x+4\right)\cdot\left(2-x\right)}=\frac{3\left(1-4x^2\right)}{2\left(-x^2-2x+8\right)}\)
\(=\frac{3-12x^2}{-2x^2-4x+16}\)
a) \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne-3;x\ne0\right)\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne0;x\ne-3\right)\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}\) \(\left(ĐKXĐ:x\ne\pm2\right)\)
\(=\frac{-5\left(x-2\right)}{2\left(x-2\right)}=\frac{-5}{2}\)