Question

giải phương trình:

\(\dfrac{2}{x-3}\) + \(\dfrac{3}{x+3}\)=\(\dfrac{7x+5}{x^2-9}\)

Answer 1

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{x^2-9}\left(x\ne3;x\ne-3\right)\\ < =>\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{\left(x-3\right)\left(x+3\right)}\)

suy ra:

`2(x+3)+3(x-3)=7x+5`

`<=>2x+6+3x-9=7x+5`

`<=>2x+3x-7x=5-6+9`

`<=> -2x=8`

`<=> x=-4(tm)`

 

Answer 2

ĐKXĐ: \(x\ne\pm3\)

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{x^2-9}\)

\(\Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{7x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2\left(x+3\right)+3\left(x-3\right)=7x+5\)

\(\Leftrightarrow2x+6+3x-9=7x+5\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\) (thỏa)

Vậy pt có nghiệm \(x=-4\)