Question

giúp mình với

 

Answer 1

\(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}=\dfrac{1}{8}\left(x\ne-1;x\ne-2;x\ne-3;x\ne-4;x\ne-5\right)\)

\(< =>\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{1}{8}\)

\(< =>\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{8}\)

\(< =>\dfrac{1}{x+1}-\dfrac{1}{x+5}=\dfrac{1}{8}\)

suy ra

`8(x+5)-8(x+1)=(x+1)(x+5)`

`<=>8x+40-8x-8=x^2 +5x+x+5`

`<=>x^2 +5x+x+5-32=0`

`<=>x^2 +6x-27=0`

`<=>(x-3)(x+9)=0`

\(< =>\left[{}\begin{matrix}x-3=0\\x+9=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=3\left(tm\right)\\x=-9\left(tm\right)\end{matrix}\right.\)