Question

[Huyền]

Giải phương trình:

a) \(\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\)

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

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{-x}{2\left(x-3\right)}\)

\(\Leftrightarrow x\left(x-3\right)-4x=-x\left(x+1\right)\)

\(\Leftrightarrow x^2-3x-4x+x^2+x=0\)

\(\Leftrightarrow2x^2-6x=0\)

=>2x(x-3)=0

=>x=0(nhận) hoặc x=3(loại)

b: \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

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

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\text{Δ}=11^2-4\cdot1\cdot\left(-26\right)=121+104=225>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-11-15}{2}=\dfrac{-26}{2}=-13\\x_2=\dfrac{-11+15}{2}=\dfrac{4}{2}=2\end{matrix}\right.\)