`1+(90-x)/(x+10)+1/9=90/x`

Đáp án: $x=-50\pm\sqrt{10600}$

Giải thích các bước giải:

ĐKXĐ: $x\ne0, -10$

Ta có:

$1+\dfrac{90-x}{x+10}+\dfrac19=\dfrac{90}{x}$

$\to 1+\dfrac19+\dfrac{90-x}{x+10}=\dfrac{90}{x}$

$\to \dfrac{10}9+\dfrac{90-x}{x+10}=\dfrac{90}{x}$

$\to 10x\left(x+10\right)+9x\left(90-x\right)=810\left(x+10\right)$

$\to x^2+910x=810x+8100$

$\to x^2+100x-8100=0$

$\to x^2+2x\cdot 50+50^2=10600$

$\to (x+50)^2=10600$

$\to x+50=\pm\sqrt{10600}$

$\to x=-50\pm\sqrt{10600}$