Question

\(\dfrac{3}{x^2+5x+4}+\dfrac{2}{x^2+10x+24}=\dfrac{4}{3}=\dfrac{9}{x^2+3x-18}\)

Answer 1

\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{9}{\left(x-3\right)\left(x+6\right)}=\dfrac{4}{3}\)

=> \(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{1}{x-3}-\dfrac{1}{x+6}=\dfrac{4}{3}\)

=> \(\dfrac{1}{x+1}-\dfrac{1}{x+6}-\dfrac{1}{x-3}+\dfrac{1}{x+6}=0\)

=> \(\dfrac{1}{x+1}-\dfrac{1}{x-3}=0\)

Ma \(\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\)

=> pt vo nghiem

Answer 2

\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}-\dfrac{1}{x+3}+\dfrac{1}{x+6}=\dfrac{4}{3}\)

=> \(\dfrac{1}{x+1}-\dfrac{1}{x+3}=\dfrac{4}{3}\)

=> \(\dfrac{2}{\left(x+1\right)\left(x+3\right)}=\dfrac{4}{3}\)

=> 4(x+1)(x+3)=6

=> 4(x²+4x+3)=6

=> 4x²+16x+6=0

=> (4x²+16x+16)-10=0

=> (2x+4)²=10

=> \(\left[{}\begin{matrix}2x+4=\sqrt{10}\\2x+4=-\sqrt{10}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-4}{2}\\x=\dfrac{-\sqrt{10}-4}{2}\end{matrix}\right.\)