Question

\(\dfrac{1}{x^2+7x+12}\) + \(\dfrac{1}{x^2+9x+20}\) + \(\dfrac{1}{x^2+11x+30}\)=\(\dfrac{1}{18}\)

Answer 1

\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+9x+18=54\)

\(\Leftrightarrow x^2+9x-36=0\)

=>(x+12)(x-3)=0

=>x=-12 hoặc x=3

Answer 2

\(ĐKXĐ:x\ne-3,-4,-5,-6\)

\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{3}{x^2+9x+18}=\dfrac{1}{18}\\ \Leftrightarrow x^2+9x+18=54\)

\(\Leftrightarrow x^2+9x-36=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-12\left(tm\right)\end{matrix}\right.\)

Answer 3

Hướng dẫn:

ĐKXĐ:...

Ta có:

\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\left(x+3\right)\left(x+6\right)=54\Leftrightarrow...\)