a) \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
( ĐKXĐ : \(x\ne0,x\ne-5\) )
\(B=\dfrac{\left(x^2+2x\right).x}{2x\left(x+5\right)}+\dfrac{\left(x-5\right).2\left(x+5\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3+2x^2+2x^2+10x-10x-50+50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3-x^2+5x^2-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^2\left(x-1\right)+5x\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{\left(x-1\right)\left(x+5\right)x}{2x\left(x+5\right)}\)
\(B=\dfrac{x-1}{2}\)
Câu b và c dễ vì đã có kết quả rút gọn rồi :)
