Giá trị của biểu thức \(P=\dfrac{10x}{x^2+3x-4}-\dfrac{2x-3}{x+4}+\dfrac{x+1}{1-x}\) tại \(x=-1\) là
\(\dfrac{7}{4}.\)\(\dfrac{4}{3}.\)\(\dfrac{10}{3}.\)\(-\dfrac{10}{3}.\)Hướng dẫn giải:ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right..\)
Ta có: \(P=\dfrac{10x}{x^2+3x-4}-\dfrac{2x-3}{x+4}+\dfrac{x+1}{1-x}\)
\(=\dfrac{10x}{\left(x+4\right)\left(x-1\right)}-\dfrac{2x-3}{x+4}-\dfrac{x+1}{x-1}\)
\(=\dfrac{10x-\left(2x-3\right)\left(x-1\right)-\left(x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{10x-2x^2+2x+3x-3-x^2-4x-x-4}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{-3x^2+10x-7}{\left(x+4\right)\left(x-1\right)}=\dfrac{-\left(x-1\right)\left(3x-7\right)}{\left(x+4\right)\left(x-1\right)}=-\dfrac{3x-7}{x+4}.\)
Với \(x=-1\) (thỏa mãn ĐKXĐ) ta có: \(P=-\dfrac{3.\left(-1\right)-7}{\left(-1\right)+4}=\dfrac{10}{3}.\)
