\(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
a) ĐKXĐ: \(x\ne-5;x\ne0\)
b) Rút gọn A:
\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\\ =\frac{x.\left(x^2+2x\right)}{2x.\left(x+5\right)}+\frac{\left(x-5\right).2.\left(x+5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\\ =\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\\ =\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=\frac{x.\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x^2+4x-5}{2.\left(x+5\right)}\)
Để A=1:
\(\Leftrightarrow\frac{x^2+4x-5}{2\left(x+5\right)}=1\\ \Leftrightarrow x^2+4x-5=2x+10\\ \Leftrightarrow x^2+4x-2x-5-10=0\\ \Leftrightarrow x^2+2x-15=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-5\left(l\right)\end{matrix}\right.\)
=> Để A=1 => x=3
