Question

\(\dfrac{3}{\left(x-4\right).\left(x-7\right)}+\dfrac{6}{\left(x-7\right).\left(x-13\right)}+\dfrac{15}{\left(x-13\right).\left(x-28\right)}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)

Answer 1

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-13}+\dfrac{1}{x-13}-\dfrac{1}{x-28}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{2}{x-28}=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x-28-2x+8}{\left(x-4\right)\left(x-28\right)}=\dfrac{-5}{2}\)

\(\Leftrightarrow-5\left(x^2-32x+112\right)=2\left(-x-20\right)\)

\(\Leftrightarrow-5x^2+160x-560=-2x-40\)

\(\Leftrightarrow-5x^2+162x-520=0\)

\(\text{Δ}=162^2-4\cdot\left(-5\right)\cdot\left(-520\right)=15844\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{162-2\sqrt{3961}}{10}\\x_2=\dfrac{162+2\sqrt{3961}}{10}\end{matrix}\right.\)