Câu 38:
\(\begin{array}{l}\left\{ \begin{array}{l}x + y = 2\\{x^2}y + x{y^2} = 4{m^2} - 2m\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + y = 2\\xy\left( {x + y} \right) = 4{m^2} - 2m\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x + y = 2\\xy.2 = 4{m^2} - 2m\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + y = 2\\xy = 2{m^2} - m\end{array} \right.\end{array}\)
Hệ có nghiệm \( \Leftrightarrow {2^2} \ge 4.\left( {2{m^2} - m} \right) \Leftrightarrow 8{m^2} - 4m \le 4 \Leftrightarrow 8{m^2} - 4m - 4 \le 0\)
\( \Leftrightarrow - \dfrac{1}{2} \le m \le 1\)
Câu 39:
\(\begin{array}{l}\sqrt {x - 1} + \dfrac{{x - m}}{{\sqrt {x - 1} }} = \dfrac{{2m}}{{\sqrt {x - 1} }}\left( {DK:x > 1} \right)\\ \Leftrightarrow \sqrt {x - 1} = \dfrac{{3m - x}}{{\sqrt {x - 1} }} \Rightarrow x - 1 = 3m - x\\ \Leftrightarrow 2x = 3m + 1 \Leftrightarrow x = \dfrac{{3m + 1}}{2}\end{array}\)
Phương trình có nghiệm \( \Leftrightarrow \dfrac{{3m + 1}}{2} > 1 \Leftrightarrow 3m + 1 > 2 \Leftrightarrow m > \dfrac{1}{3}\)
