\(2,\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)
\(Đkxđ:x\ne1;x\ne-2\)
\(\Leftrightarrow\left(3x+2\right)\left(x+2\right)+\left(2x-4\right)\left(x-1\right)=5\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow x=6\)
Vậy ...........
\(1.5\left(x^2-2x\right)=\left(3+5x\right)\left(x-1\right)\\\Leftrightarrow 5x^2-10x=3x-3+5x^2-5x\\ \Leftrightarrow5x^2-5x^2-10x-3x+5x=-3\\ \Leftrightarrow-8x=-3\\\Leftrightarrow x=\frac{3}{8}\)
Vậy nghiệm của phương trình trên là \(\frac{3}{8}\)
1: Ta có: \(5\left(x^2-2x\right)=\left(3+5x\right)\left(x-1\right)\)
\(\Leftrightarrow5x^2-10x=3x-3+5x^2-5\)
\(\Leftrightarrow5x^2-10x-3x+3-5x^2+5=0\)
\(\Leftrightarrow-13x+8=0\)
\(\Leftrightarrow-13x=-8\)
\(\Leftrightarrow13x=8\)
\(\Leftrightarrow x=\frac{8}{13}\)
Vậy: \(x=\frac{8}{13}\)
2) ĐKXĐ: \(x\ne1;x\ne-2\)
Ta có: \(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)
\(\Leftrightarrow\frac{3x+2}{x-1}+\frac{2x-4}{x+2}-5=0\)
\(\Leftrightarrow\frac{\left(3x+2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{\left(2x-4\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{5\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+2\right)+\left(2x-4\right)\left(x-1\right)-5\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow3x^2+6x+2x+4+2x^2-2x-4x+4-\left(5x+10\right)\left(x-1\right)=0\)
\(\Leftrightarrow5x^2+2x+8-\left(5x^2-5x+10x-10\right)=0\)
\(\Leftrightarrow5x^2+2x+8-5x^2+5x-10x+10=0\)
\(\Leftrightarrow-3x+18=0\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)(thỏa mãn điều kiện)
Vậy: x=6
