\(ĐKXĐ:x\ne3;x\ne-1;x\ne0\)
a)\(P=\frac{x^2-4x+3}{\left(x+1\right)\left(x-3\right)}:\left(x+1-\frac{1}{x+1}\right)\)
\(\Leftrightarrow P=\frac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-3\right)}:\frac{\left(x+1\right)^2-1}{x+1}\)
\(\Leftrightarrow P=\frac{x-1}{x+1}:\frac{x^2+2x+1-1}{x+1}\)
\(\Leftrightarrow P=\frac{x-1}{x+1}.\frac{x+1}{x^2+2x}\)
\(\Leftrightarrow P=\frac{x-1}{x^2+2x}\)
b) Để \(P=\frac{2}{15}\)
\(\Leftrightarrow\frac{x-1}{x^2+2x}=\frac{2}{15}\)
\(\Leftrightarrow15\left(x-1\right)=2\left(x^2+2x\right)\)
\(\Leftrightarrow15x-15=2x^2+4x\)
\(\Leftrightarrow2x^2+4x-15x+15=0\)
\(\Leftrightarrow2x^2-11x+15=0\)
\(\Leftrightarrow2x^2-6x-5x+15=0\)
\(\Leftrightarrow2x\left(x-3\right)-5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=\frac{5}{2}\left(tm\right)\end{matrix}\right.\)
Vậy để \(P=\frac{2}{15}\Leftrightarrow x=\frac{5}{2}\)
