ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x+m=2\sqrt{x-1}\Leftrightarrow x^2+2mx+m^2=4\left(x-1\right)\)
\(\Leftrightarrow x^2+2\left(m-2\right)x+m^2+4=0\) (1)
a/ Bạn tự giải
b/ Để phương trình có 2 nghiệm pb \(\Leftrightarrow\left(1\right)\) có 2 nghiệm pb thỏa mãn:
\(1\le x_1< x_2\Leftrightarrow\left\{{}\begin{matrix}\Delta'=\left(m-2\right)^2-\left(m^2+4\right)>0\\f\left(1\right)=1^2+2\left(m-2\right)+m^2+4\ge0\\\frac{S}{2}=\frac{-2\left(m-2\right)}{2}>1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-4m>0\\m^2+2m+1\ge0\\2-m>1\end{matrix}\right.\) \(\Rightarrow m< 0\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-2\right)\\x_1x_2=m^2+4\end{matrix}\right.\)
\(x_1+x_2=x_1x_2+\frac{3}{4}\)
\(\Leftrightarrow-2\left(m-2\right)=m^2+4+\frac{3}{4}\)
\(\Leftrightarrow m^2+2m+\frac{3}{4}=0\Rightarrow\left[{}\begin{matrix}m=-\frac{1}{2}\\m=-\frac{3}{2}\end{matrix}\right.\)
