Question

[Rita Hương Rika]

GIAI GIÚP EM VỚI

Giải pt chứa an ở mẫu sau

2x/x-3+x/x+3=2x^2/x^2-9

em sap thi rui ạ

Answer 1

[Gia Hân Ngô]

\(\dfrac{2x}{x-3}+\dfrac{x}{x+3}=\dfrac{2x^2}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

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

\(\Rightarrow2x\left(x+3\right)+x\left(x-3\right)=2x^2\)

\(\Leftrightarrow2x^2+6x+x^2-3x-2x^2=0\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy ............................

Answer 2

[Nhã Doanh]

ĐKXĐ: x khác 3 và x khác -3

\(\dfrac{2x}{x-3}+\dfrac{x}{x+3}=\dfrac{2x^2}{x^2-9}\)

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

\(\Leftrightarrow2x^2+6x+x^2-3x=2x^2\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy......