Question

[Hiên Mai]

\(\frac{2x+1}{2x^2+5x+2}\)\(-\frac{3}{x^2-4}\)\(=2\)

giải phương trình sau:

Answer 1

[Nguyễn Minh Đăng]

đk: \(x\ne\left\{\pm2;-\frac{1}{2}\right\}\)

Ta có: \(\frac{2x+1}{2x^2+5x+2}-\frac{3}{x^2-4}=2\)

\(\Leftrightarrow\frac{\left(2x+1\right)\left(x-2\right)-3\left(2x+1\right)}{\left(x+2\right)\left(x-2\right)\left(2x+1\right)}=2\)

\(\Leftrightarrow2x^2-3x-2-6x-3=2\left(x^2-4\right)\left(2x+1\right)\)

\(\Leftrightarrow2x^2-9x-5=4x^3+2x^2-16x-8\)

\(\Leftrightarrow4x^3-7x-3=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)\left(x+1\right)=0\)

\(\Rightarrow x\in\left\{\frac{3}{2};-1\right\}\) (đkxđ)

Answer 2

[Tuấn Anh]

\(\frac{2x+1}{2x^2+5x+2}-\frac{3}{x^2-4}=2\left(x\ne-\frac{1}{2};\pm2\right)\)

\(\Leftrightarrow\frac{2x+1}{2x\left(x+2\right)+\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=2\)

\(\Leftrightarrow\frac{2x+1}{\left(2x+1\right)\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{x-2}{\left(x-2\right)\left(x+2\right)}-\frac{3}{\left(x-2\right)\left(x+2\right)}=\frac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x-2-3=2x^2-8\)

\(\Leftrightarrow x-2x^2=-8+2+3\)

\(\Leftrightarrow x-2x^2=-3\)

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

\(\Leftrightarrow2x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\left(tm\right)\\x=-1\left(tm\right)\end{cases}}\)

Vậy nghiệm của phương trình là: \(S=\left\{\frac{3}{2};-1\right\}\)