\(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)
\(\frac{\left(x+2\right)\left(x-2\right)+3\left(x+1\right)}{x^2-x-2}=\frac{3+x^2-x-2}{x^2-x-2}\)
\(x^2-4+3x+3=1+x^2-x\)
\(x^2+3x-1-1-x^2+x=0\)
\(4x-2=0\)
\(4x=2\Leftrightarrow x=\frac{1}{2}\)
Vậy.....
\(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)
\(\Leftrightarrow\)\(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{\left(x+1\right).\left(x-2\right)}+1\)
ĐKXĐ: \(x\ne-1,2\)
\(\frac{\left(x+2\right).\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}+\)\(\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x-2\right)}=\)\(\frac{3}{\left(x+1\right).\left(x-2\right)}+\frac{\left(x+1\right).\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}\)
\(\Leftrightarrow\) \(\left(x^2-4\right)\) \(+3.\left(x+1\right)=\)\(3+\left(x+1\right).\left(x-2\right)\)
\(\Leftrightarrow\) x2 - 4 + 3x + 3 = 3 + x2 - x - 2
\(\Leftrightarrow\) x2 + 3x - x2 + x = 4 - 3 + 3 - 2
\(\Leftrightarrow\) 4x = 2
\(\Leftrightarrow\)\(x=\frac{1}{2}\)
Vậy phương trình có nghiệm là: \(x=\frac{1}{2}\)
\(\frac{\left(x+2\right)\left(x-2\right)+3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3+x^2-x-2}{x^2-x-2}\)
\(\frac{x^2-2^2+3x+3}{x^2-2x+x-2}=\frac{3+x^2-x-2}{x^2-x-2}\)
\(\frac{x^2+3x-1-3-x^2+x+2}{x^2-x-2}=0\)
\(\frac{4x-3}{x^2-x-2}=0\)
\(\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)
cho tớ sửa đoạn \(\frac{x^2+3x-1-3-x^2+x+2}{x^2-x-2}=0\)
\(\frac{4x-2}{x^2-x-2}=0\Leftrightarrow4x-2=0\Leftrightarrow x=\frac{2}{4}=\frac{1}{2}\)
