Ta có: \(\Delta=\left(-2\right)^2-4\cdot1\cdot\left(-m+1\right)\)
=4+4m-4
=4m
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=2\\ x_1x_2=\frac{c}{a}=-m+1\end{cases}\)
Để phương trình có hai nghiệm dương thì \(\begin{cases}\Delta>=0\\ x_1+x_2>0\\ x_1x_2>0\end{cases}\Rightarrow\begin{cases}4m\ge0\\ 2>0\\ -m+1>0\end{cases}\Rightarrow\begin{cases}m\ge0\\ -m>-1\end{cases}\Rightarrow0\le m<1\)
\(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}=2\)
=>\(\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)^2=2^2=4\)
=>\(\frac{1}{x_1}+\frac{1}{x_2}+2\cdot\sqrt{\frac{1}{x_1x_2}}=4\)
=>\(\frac{x_1+x_2}{x_1x_2}+\frac{2}{\sqrt{x_1x_2}}=4\)
=>\(\frac{2}{x_1x_2}+\frac{2}{\sqrt{x_1x_2}}=4\)
=>\(\frac{1}{x_1x_2}+\frac{1}{\sqrt{x_1x_2}}=2\)
=>\(\frac{1}{-m+1}+\frac{1}{\sqrt{-m+1}}=2\)
=>\(\frac{1}{-m+1}+\frac{\sqrt{-m+1}}{-m+1}=2\)
=>\(2\cdot\left(-m+1\right)=\sqrt{-m+1}+1\)
=>\(2\cdot\sqrt{-m+1}^2-\sqrt{-m+1}-1=0\)
=>\(\left(\sqrt{-m+1}-1\right)\left(2\cdot\sqrt{-m+1}+1\right)=0\)
=>\(\sqrt{-m+1}-1=0\)
=>-m+1=1
=>-m=0
=>m=0(nhận)
