Bài 1:
a) Thay x=-1 vào B, ta được:
\(B=\dfrac{-5}{-1-4}=\dfrac{-5}{-5}=1\)
Vậy: Khi x=-1 thì B=1
Bài 1:
b) Ta có: \(A=\dfrac{x+3}{x+5}-\dfrac{-x+9}{\left(x+5\right)\left(x-2\right)}+\dfrac{1}{2-x}\)
\(=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}+\dfrac{x-9}{\left(x+5\right)\left(x-2\right)}-\dfrac{x+5}{\left(x-2\right)\left(x+5\right)}\)
\(=\dfrac{x^2-2x+3x-6+x-9-x-5}{\left(x+5\right)\cdot\left(x-2\right)}\)
\(=\dfrac{x^2+x-20}{\left(x+5\right)\left(x-2\right)}\)
\(=\dfrac{\left(x+5\right)\left(x-4\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
Bài 1:
c) Ta có: P=AB
nên \(P=\dfrac{x-4}{x-2}\cdot\dfrac{-5}{x-4}=\dfrac{-5}{x-2}\)
Để P<0 thì x-2>0
hay x>2
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)
