Giải thích các bước giải:
Bài 2:
a.Với $m=\sqrt2$
$\to\begin{cases}\sqrt{2}x-y=2\\3x+y\sqrt{2}=5\end{cases}$
$\to\begin{cases}y=\sqrt{2}x-2\\3x+(\sqrt{2}x-2)\sqrt{2}=5\end{cases}$
$\to\begin{cases}y=\sqrt{2}x-2\\3x+2x-2\sqrt{2}=5\end{cases}$
$\to\begin{cases}y=\sqrt{2}x-2\\5x=2\sqrt{2}+5\end{cases}$
$\to\begin{cases}y=\sqrt{2}.\dfrac{2\sqrt{2}+5}{5}-2=\dfrac{-6+5\sqrt{2}}{5}\\x=\dfrac{2\sqrt{2}+5}{5}\end{cases}$
b.Ta có : $mx-y=2\to y=mx-2$
$\to 3x+m(mx-2)=5$
$\to x(m^2+3)=2m+5$
$\to x=\dfrac{2m+5}{m^2+3}, m^2+3>0$
$\to y=m.\dfrac{2m+5}{m^2+3}-2=\dfrac{5m-6}{m^2+3}$
Để $x+y<1\to \dfrac{2m+5}{m^2+3}+\dfrac{5m-6}{m^2+3}<1$
$\to \dfrac{2m+5+5m-6}{m^2+3}<1$
$\to \dfrac{7m-1}{m^2+3}<1$
$\to 7m-1<m^2+3(m^2+3>0)$
$\to m^2-7m+4>0$
$\to \left(m-\dfrac{7}{2}\right)^2-\dfrac{33}{4}>0$
$\to \left(m-\dfrac{7}{2}\right)^2>\dfrac{33}{4}$
$\to m<\dfrac{-\sqrt{33}+7}{2}\quad \mathrm{or}\quad \:m>\dfrac{\sqrt{33}+7}{2}$
