\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\) ;
(ĐKXĐ: x ≠ 0; x ≠ \(-1\); x ≠ 4; x ≠ \(-2\))
\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\)
\(=\dfrac{x+1}{x^2-4x+2x-8}.\dfrac{4-x}{x\left(x+1\right)}\)
\(=\dfrac{x+1}{x\left(x-4\right)+2\left(x-4\right)}.\dfrac{4-x}{x\left(x+1\right)}\)
\(=\dfrac{x+1}{\left(x-4\right)\left(x+2\right)}.\dfrac{4-x}{x\left(x+1\right)}\)
\(=\dfrac{4-x}{x\left(x-4\right)\left(x+2\right)}\)
\(=\dfrac{-\left(x-4\right)}{-x\left(x-4\right)\left(x+2\right)}\)
\(=\dfrac{-1}{-x\left(x+2\right)}\)
\(=\dfrac{1}{x\left(x+2\right)}\)
dạ, vâng ạ:
-(x+1)(x-4) / ((x-1)2 - 32)x(x+1) = - 1/ x(x+2)
bài trước mình làm sai rồi. mình sửa lại nè
\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\);
( ĐKXĐ: x ≠ 0; x ≠ \(-1\); x ≠ 4; x ≠ \(-2\) )
\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\)
\(=\dfrac{x+1}{x^2-4x+2x-8}.\dfrac{4-x}{x\left(x+1\right)}\)
\(=\dfrac{x+1}{x\left(x-4\right)+2\left(x-4\right)}.\dfrac{4-x}{x\left(x+1\right)}\)
\(=\dfrac{x+1}{\left(x-4\right)\left(x+2\right)}.\dfrac{4-x}{x\left(x+1\right)}\)
\(=\dfrac{4-x}{x\left(x-4\right)\left(x+2\right)}\)
=\(\dfrac{-\left(x-4\right)}{x\left(x-4\right)\left(x+2\right)}\)
\(=\dfrac{-1}{x\left(x+2\right)}\)
