Lời giải:
$C=\frac{x^2-2x+1}{2x^2-x-1}=\frac{(x-1)^2}{(x-1)(2x+1)}$
ĐKXĐ: $x\neq 1; x\neq \frac{-1}{2}$
Thu gọn $C=\frac{x-1}{2x+1}$
Với $x^2-5x+4=0$
$\Leftrightarrow (x-1)(x-4)=0$
Do $x-1\neq 0$ theo ĐKXĐ nên $x-4=0$
$\Leftrightarrow x=4$
Khi đó: $C=\frac{4-1}{2.4+1}=\frac{3}{9}=\frac{1}{3}$
Ta có:
\(x^2-5x+4=0\)
\(\Leftrightarrow x^2-4x-x+4=0\)
\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
\(C=\dfrac{x^2-2x+1}{2x^2-x-1}\left(x\ne1;x\ne-\dfrac{1}{2}\right)\)
\(C=\dfrac{x^2-2\cdot x\cdot1+1^2}{2x^2-2x+x-1}=\dfrac{\left(x-1\right)^2}{2x\left(x-1\right)+\left(x-1\right)}\)
\(C=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)
Nên x = 1 là không thỏa mãn
Thay x = 4 vào C ta có:
\(C=\dfrac{4-1}{2\cdot4+1}=\dfrac{3}{9}=\dfrac{1}{3}\)
