Đáp án:
$\begin{array}{l}
b)Dkxd:\left\{ \begin{array}{l}
{x^2} - 5x + 4 \ge 0\\
2x - 2 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 4\\
x \le 1
\end{array} \right.\\
x \ge 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \ge 4\\
x = 1
\end{array} \right.\\
\sqrt {{x^2} - 5x + 4} \le 2x - 2\\
\Rightarrow {x^2} - 5x + 4 \le 4{x^2} - 8x + 4\\
\Rightarrow 3{x^2} - 3x \ge 0\\
\Rightarrow 3x\left( {x - 1} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le 0
\end{array} \right.\\
Vậy\,x = 1\,hoặc\,x \ge 4
\end{array}$
Bài 2: f(x) >0 vô nghiệm
$\begin{array}{l}
\Rightarrow f\left( x \right) \le 0\forall x\\
\Rightarrow \left\{ \begin{array}{l}
m - 1 = 0\\
- 2.\left( {m - 1} \right) = 0\\
- 1 < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
m = 1\\
- 1 < 0
\end{array} \right. \Rightarrow m = 1\\
Hoặc:\left\{ \begin{array}{l}
m - 1 < 0\\
\Delta ' \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < 1\\
{\left( {m - 1} \right)^2} + m - 1 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < 1\\
\left( {m - 1} \right)\left( {m - 1 + 1} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < 1\\
0 \le m \le 1
\end{array} \right.\\
Vậy\,0 \le m \le 1
\end{array}$
