bt em gửi cô Thương
1)\(ĐKXĐ\hept{\begin{cases}x\ne1\\x\ne3\end{cases}}\)
\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-4x+3}=0\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-x-3x+3}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{8}{x\left(x-1\right)-3\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\frac{x^2-1}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{2x-6}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)( tm)
Vậy nghiemj của pt x=3
2)\(x^3-x^2-9x+9=0\)
\(\Leftrightarrow x^2\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)hoặc x+3=0
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)hoặc x=-3
Vậy tập hợp nghiệm \(S=\left\{1;3;-3\right\}\)
