Bài làm
Ta có: \(\frac{x+a}{x+3}+\frac{x-3}{x-a}=2\)
Thay a = 2 vào phương trình trên, ta được:
\(\frac{x+2}{x+3}+\frac{x-3}{x-2}=2\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=2\)
\(\Leftrightarrow\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}+\frac{x^2-9}{\left(x-2\right)\left(x+3\right)}=2\)
\(\Leftrightarrow\frac{x^2-4+x^2-9}{\left(x+3\right)\left(x-2\right)}=2\)
\(\Leftrightarrow\frac{2x^2-13}{\left(x+3\right)\left(x-2\right)}-2=0\)
\(\Leftrightarrow\frac{2\left(2x^2-13\right)}{2\left(x+3\right)\left(x-2\right)}-\frac{4\left(x+3\right)\left(x-2\right)}{2\left(x+3\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{4x^2-26}{2\left(x+3\right)\left(x-2\right)}-\frac{4\left(x^2-x-6\right)}{2\left(x+3\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{4x^2-26-4x^2+4x+24}{2\left(x+3\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{4x-2}{2\left(x+3\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{2\left(x+3\right)\left(x-2\right)}=0\)
\(\Leftrightarrow2\left(2x-1\right):2\left(x+3\right)\left(x-2\right)=0\)4
\(\Leftrightarrow2\left(2x-1\right)=0.2\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(2x-1\right)=0\)
Mà 2 > 0
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy tập nghiệm của phương trình trên là S = { 1/2 } khi a = 2
# Học tốt #
