\(\frac{4}{x-5}\)- \(\frac{1}{x+5}\)+\(\frac{13x-x^2}{\left(5-x\right)\left(5+x\right)}\)= \(\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)- \(\frac{x-5}{\left(x-5\right)\left(x+5\right)}\)- \(\frac{x\left(x-13\right)}{\left(x-5\right)\left(5+x\right)}\)=\(\frac{4\left(x+5\right)-x+5+x\left(x-13\right)}{\left(x-5\right)\left(x+5\right)}\)= \(\frac{x^2-10x+12}{\left(x-5\right)\left(x+5\right)}\)

\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{\left(5-x\right)\left(x+5\right)}\)ĐKXĐ : \(x\ne\pm5\)

\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x+5\right)\left(x-5\right)}-\frac{13x-x^2}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{4x+20-x+5-13x+x^2}{\left(x-5\right)\left(x+5\right)}=\frac{-10x+25+x^2}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)

\(\Leftrightarrow\)\(\frac{x+1}{x-2}+\frac{8-7x}{x^2-2x}+\frac{2}{x}\left(ĐKXĐ.x\ne2.x\ne0\right)\)

\(\Leftrightarrow\)\(\frac{x\left(x+1\right)}{x\left(x-2\right)}+\frac{8-7x}{x\left(x-2\right)}+\frac{2\left(x-2\right)}{x\left(x-2\right)}\)

\(\Leftrightarrow\frac{x^2+x+8-7x+2x-4}{x\left(x-2\right)}\)

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

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

b, lấy cái dễ nhất làm cho nhanh :PP làm tương tự thôi nhé !!!

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

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

\(=\frac{4x-4-x^2+x-x^2-5x}{MTC}=\frac{-4-2x^2}{MTC}\)

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

\(\frac{x-1}{x+2}-\frac{x+1}{2-x}-\frac{x^2-2x+4}{x^2-4}\)

= \(\frac{x-1}{x+2}+\frac{x+1}{x-2}-\frac{x^2-2x+4}{x^2-4}\)

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

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

im,em moi lop 3m :)