a: Ta có: x2+5x=0
=>x(x+5)=0
=>x=0(nhận) hoặc x=-5(loại)
Thay x=0 vào A, ta được:
\(A=1-\dfrac{8}{0+5}=1-\dfrac{8}{5}=-\dfrac{3}{5}\)
b: \(B=\dfrac{\left(x+3\right)^2-\left(x-3\right)^2+6-10x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+6x+9-x^2+6x-9+6-10x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\)
c: Để A=-5/3 thì \(1-\dfrac{8}{x+5}=\dfrac{-5}{3}\)
\(\Leftrightarrow\dfrac{x+5-8}{x+5}=\dfrac{-5}{3}\)
\(\Leftrightarrow\dfrac{x-3}{x+5}=\dfrac{-5}{3}\)
=>-5x-25=3x-9
=>-8x=16
hay x=-2(nhận)
Đúng 0
Bình luận (0)
