Sửa đề: \(x^2-2x+m+1=0\)
\(\text{Δ}=\left(-2\right)^2-4\cdot1\cdot\left(m+1\right)=4-4m-4=-4m\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>-4m>0
=>m<0
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=m+1\end{matrix}\right.\)
\(x_1+x_2=2\)
=>\(x_2+\dfrac{1}{\sqrt{x_2}}=2\)
=>\(x_2\sqrt{x_2}+1=2\sqrt{x_2}\)
\(\Leftrightarrow\left(\sqrt{x_2}\right)^3-2\sqrt{x_2}+1=0\)
=>\(\left(\sqrt{x_2}\right)^3-x_2+x_2-\sqrt{x_2}-\sqrt{x_2}+1=0\)
=>\(\left(\sqrt{x_2}-1\right)\left(x_2+\sqrt{x_2}-1\right)=0\)
=>\(\left[{}\begin{matrix}x_2=1\left(nhận\right)\\\sqrt{x_2}=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x_2=1\\x_2=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1}{1}=1\\x_1=\dfrac{1}{\sqrt{\dfrac{3-\sqrt{5}}{2}}}=\dfrac{1}{\sqrt{\dfrac{6-2\sqrt{5}}{4}}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=1\\x_1=1:\dfrac{\sqrt{5}-1}{2}=\dfrac{2}{\sqrt{5}-1}=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
TH1: x2=1; x1=1
\(x_1\cdot x_2=m+1\)
=>\(m+1=1\cdot1=1\)
=>m=0(loại)
TH2: \(x_2=\dfrac{3-\sqrt{5}}{2};x_1=\dfrac{1+\sqrt{5}}{2}\)
\(x_1x_2=m+1\)
=>\(m+1=\dfrac{\left(3-\sqrt{5}\right)\left(1+\sqrt{5}\right)}{4}=\dfrac{-2+2\sqrt{5}}{4}=\dfrac{\sqrt{5}-1}{2}\)
=>\(m=\dfrac{\sqrt{5}-1}{2}-1=\dfrac{\sqrt{5}-3}{2}\)(nhận)
