\(ĐKXĐ:x\ne-3;x\ne2;x\ne-1;x\ne\frac{1}{2}\)
Xét\(VT=\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}\)
\(=\frac{5\left(x+1\right)}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}-\frac{2\left(x-2\right)}{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)
\(=\frac{5x+5-2x+4}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}\)
\(=\frac{3x+9}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}=\frac{3}{\left(x-2\right)\left(x+1\right)}\)
\(pt\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{3}{4x-2}\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=4x-2\)
\(\Leftrightarrow x^2-x-2=4x-2\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)(tm)
Vậy tập nghiệm của phương trình là {0;5}
ĐKXĐ: \(x\ne-3,2,-1\)
\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4x+3}=\frac{3}{4x-2}\)
\(\Leftrightarrow\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{3}{2\left(x-2\right)}\)
\(\Leftrightarrow10\left(x+1\right)\left(2x-1\right)-4\left(x-2\right)\left(2x-1\right)=3\left(x-2\right)\left(x+3\right)\left(x+1\right)\)
\(\Leftrightarrow12x^2+30x-18=3x^2+6x^2-15x-18\)
\(\Leftrightarrow12x^2+30x=3x^3+6x^2-15\)
\(\Leftrightarrow12x^2+30x-3x^3-6x^2+15x=0\)
\(\Leftrightarrow6x^2+45x-3x^2=0\)
\(\Leftrightarrow3x\left(2x+15-x^2\right)=0\)
\(\Leftrightarrow-x\left(x^2-2x-15\right)=0\)
\(\Leftrightarrow-x\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}-x=0\\x-5=0\\x+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(tm\right)\\x=5\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)
Vậy: tập nghiệm của phương trình là: S = {0, 5}
