Question

Giải các phương trình sau:

 \(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

 

Answer 1

Ta có: \(\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)

\(\Rightarrow\frac{6-x^2-2}{x^2+2}+\frac{12-x^2-8}{x^2+8}=\frac{x^2+3-7}{x^2+3}\)

\(\Rightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+8}-\frac{x^2-4}{x^2+3}=0\)

\(\Rightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+3}=0\)

\(\Rightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)

mà \(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\)

=> 4 - x² = 0 => x² = 4 => x = 2 hoặc x = -2

                                                               Vậy x = 2, x = -2