Đặt \(A=\frac{-x^2+x-11}{x^2-2x+1}\)
\(=\frac{-x^2+2x-1-x-10}{x^2-2x+1}\)
\(=-1+\frac{-x-10}{\left(x-1\right)^2}=-1+\frac{-x+1-11}{\left(x-1\right)^2}=-1-\frac{1}{\left(x-1\right)}-\frac{11}{\left(x-1\right)^2}\)
\(=-11\left\lbrack\frac{1}{\left(x-1\right)^2}+\frac{1}{11}\cdot\frac{1}{x-1}+\frac{1}{11}\right\rbrack\)
\(=-11\left\lbrack\frac{1}{\left(x-1\right)^2}+2\cdot\frac{1}{x-1}\cdot\frac{1}{22}+\frac{1}{484}+\frac{43}{484}\right\rbrack=-11\left\lbrack\left(\frac{1}{x-1}+\frac{1}{11}\right)^2\right\rbrack-\frac{43}{44}\le-\frac{43}{44}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\frac{1}{x-1}+\frac{1}{11}=0\)
=>\(\frac{1}{x-1}=\frac{1}{-11}\)
=>x-1=-11
=>x=-11+1=-10(nhận)
