d) \(\frac{2x+3}{5x-3}-\frac{3}{4x-6}=\frac{2}{5}\)
ĐKXĐ: x\(\ne\)0
Ta có:
\(\frac{2x+3}{5x-3}-\frac{3}{2\left(2x-3\right)}=\frac{2}{5}\)
\(\Rightarrow\) \(\frac{\left(2x+3\right)10\left(2x-3\right)}{10\left(5x-3\right)\left(2x-3\right)}-\frac{15\left(5x-3\right)}{10\left(5x-3\right)\left(2x-3\right)}=\frac{4\left(2x-3\right)\left(5x-3\right)}{10\left(2x-3\right)\left(5x-3\right)}\)
\(\Leftrightarrow\) \(40x^2-90-75x+45=4\left(10x^2-21x+9\right)\)
\(\Leftrightarrow\) \(40x^2-90-75x+45=40x^2-84x+36\)
\(\Leftrightarrow\) \(40x^2-40x^2-75x+84x=90-45+36\)
\(\Leftrightarrow\) \(9x=81\) \(\Rightarrow\) \(x=9\) (nhận)
Vậy PT có tập nghiệm S=\(\left\{9\right\}\)
e) ĐKXĐ: \(x\)\(\ne\)\(\pm\)1
Ta có:
\(\frac{2}{x^2+2x+1}-\frac{5}{x^2-2x+1}=\frac{3}{1-x^2}\)
\(\Rightarrow\) \(\frac{2}{\left(x+1\right)^2}-\frac{5}{\left(x-1\right)^2}=\frac{3}{\left(1-x\right)\left(1+x\right)}\)
\(\Rightarrow\)\(\frac{2\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\)- \(\frac{5\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}\)= \(\frac{3\left(x^2-1\right)}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(\Leftrightarrow\)\(2\left(x^2-2x+1\right)\)- \(5\left(x^2+2x+1\right)\)= \(3x^2-3\)
\(\Leftrightarrow\) \(2x^2-4x+2\)- \(5x^2-10x-5\)= \(3x^2-3\)
\(\Leftrightarrow\) \(2x^2-5x^2+3x^2\)- \(4x-10x\)= \(5-2-3\)
\(\Leftrightarrow\) \(14x=0\)
\(\Rightarrow\) \(x=0\)(nhận)
Vậy PT có tập nghiệm S=0
