Question

Tìm x : \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

Answer 1

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\) (ĐKXĐ: \(x\notin\left\{-4;-5;-6;-7\right\}\))

<=> \(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}-\dfrac{1}{18}=0\)

<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)

<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)

<=> \(\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}=0\)

=> \(18\left(x+7\right)-18\left(x+4\right)-\left(x+4\right)\left(x+7\right)=0\)

<=> 18x + 18.7 - 18x - 18.4 - x² - 7x - 4x - 28 = 0

<=> - x² - 11x + 26 = 0

<=> (x - 2)(x + 13) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\) (nhận)

Vậy S = {-13; 2}