\(a,đk:\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\\ b,P=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x^2+10x}\\ =\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{x\left(2x+10\right)}\\ =\dfrac{x.\left(x^2+2x\right)+\left(x-5\right).\left(2x+10\right)+50-5x}{x\left(2x+10\right)}\\ =\dfrac{x^3+2x^2+2x^2-50+50-5x}{x\left(2x+10\right)}\\ =\dfrac{x^3+4x^2-5x}{x\left(2x+10\right)}=\dfrac{x\left(x^2+4x-5\right)}{x\left(2x+10\right)}=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}=\dfrac{\left(x-1\right)\left(x+5\right)}{2\left(x+5\right)}=\dfrac{x-1}{2}\)
\(c,\) \(x=-5\left(kot/mđk\right)\)
Vậy tại \(x=-5\) P không xác định
a: ĐKXĐ: x<>0; x<>-5
b: \(P=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
c: Khi x=-5 thì P ko có giá trị
\(P=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x^2+10x}\left(dkxd:x\ne-5;0\right)\)
\(=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x^2\left(x+2\right)+2\left(x-5\right)\left(x+5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x-1}{2}\)
\(x=-5\left(ktm\right)\)
