a)
ĐKXĐ của A:
\(x^2-25\ne0\Leftrightarrow x^2\ne25\Leftrightarrow x\ne\pm5\)
b)
Ta có:
\(A=\dfrac{2x}{x^2-25}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\left(\cdot\right)\)
c) Ta có:
\(x^2+5x=0\Rightarrow x\left(x+5\right)=0\Rightarrow x\in\left\{0;-5\right\}\)
Đối chiếu đkxđ có x=0( TM)
Thay vào (.) có:
\(A-\dfrac{4}{0-5}=\dfrac{4}{5}\)
a.
A xác định \(\Leftrightarrow x^2-25\ne0\Leftrightarrow x\ne5;x\ne-5\)
b.
\(A=\dfrac{2x}{x^2-25}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\\ =\dfrac{2x}{x^2-25}-\dfrac{5\left(x+5\right)}{x^2-25}-\dfrac{x-5}{x^2-25}\\ =\dfrac{2x-5x-25-x+5}{x^2-25}\\ =\dfrac{-4x-20}{x^2-25}\\ =\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\)
c.
\(x^2+5x=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
x=0 => A = 4/5
x=-5 => A = 2/5
