Ta có:
\(8x^2-8x+m+1=0\left(a=8;b'=-4;c=m+1\right)\)
Xét \(\Delta'=16-8m-8=8-8m\)
để pt có nghiệm \(\Leftrightarrow\Delta'\ge0\Leftrightarrow8-8m\ge0\Leftrightarrow m\le1\)
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{4+\sqrt{8-8m}}{8}=\frac{4+2\sqrt{2-2m}}{8}=\frac{2+\sqrt{2-2m}}{4}\)
Vì \(x_1=\frac{1}{2}\Rightarrow\frac{2+\sqrt{2-2m}}{4}=\frac{1}{2}\)
\(\Rightarrow2+\sqrt{2-2m}=2\)
\(\Leftrightarrow2-2m=0\)
\(\Leftrightarrow m=1\)(tm đk)
Vì \(m=1\Rightarrow\Delta'=0\Rightarrow\)pt có nghiệm kép\(\Rightarrow x_1=x_2=\frac{1}{2}\)
