Câu hỏi của Nguyễn Thị Bình Yên

Question

1/Tìm x

$\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0$

2/ Cho $\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1$

Tính S = $\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}$

Nguyễn Nam · Accepted Answer

1) $\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0$

$\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=0$

$\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}=0$

$\Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow\dfrac{x^2+2.x.3+3^2}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow\dfrac{x+3}{x-3}=0$

$\Leftrightarrow x+3=0$

$\Leftrightarrow x=-3$

Vậy x=-3Câu trả lời: 1) $\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0$