Đáp án: a.$M=\dfrac{x+5}{5}$ $b.x=-\dfrac{35}9$
Giải thích các bước giải:
a.Ta có:
$M=\dfrac{x^2}{5x+25}+\dfrac{2(x-5)}{x}+\dfrac{50+5x}{x(x+5)}$
$\to M=\dfrac{x^2}{5(x+5)}+\dfrac{2(x-5)}{x}+\dfrac{50+5x}{x(x+5)}$
$\to M=\dfrac{x^3}{5x(x+5)}+\dfrac{10(x+5)(x-5)}{5x(x+5)}+\dfrac{250+25x}{5x(x+5)}$
$\to M=\dfrac{x^3+10(x+5)(x-5)+250+25x}{5x(x+5)}$
$\to M=\dfrac{x^3+10(x^2-5^2)+250+25x}{5x(x+5)}$
$\to M=\dfrac{x^3+10x^2+25x}{5x(x+5)}$
$\to M=\dfrac{x(x^2+10x+25)}{5x(x+5)}$
$\to M=\dfrac{x(x+5)^2}{5x(x+5)}$
$\to M=\dfrac{x+5}{5}$
b.Để $M=\dfrac29$
$\to\dfrac{x+5}{5}=\dfrac29$
$\to x+5=\dfrac{10}9$
$\to x=-\dfrac{35}9$
