1) \(\Delta'=1^2-\left(m-1\right)=2-m\)
Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)
Khi đó \(x_1=1+\sqrt{2-m};x_2=1-\sqrt{2-m}\)
TH1: \(2\left(1+\sqrt{2-m}\right)-\left(1-\sqrt{2-m}\right)=7\Leftrightarrow1+3\sqrt{2-m}=7\)
\(\Leftrightarrow\sqrt{2-m}=2\Leftrightarrow2-m=4\Rightarrow m=-2\left(tm\right)\)
TH2: \(2\left(1-\sqrt{2-m}\right)-\left(1+\sqrt{2-m}\right)=7\Leftrightarrow1-3\sqrt{2-m}=7\) (VÔ LÝ)
Vậy m = - 2.
2) \(P=\frac{x^4+3x^2+1}{x^2+1}=\frac{\left(x^4+2x^2+1\right)+\left(x^2+1\right)+2}{x^2+1}=\left(x^2+1\right)+\frac{2}{x^2+1}+1\)
Vì \(x^2+1\ge1\), áp dụng bđt Cô si ta có:
\(\left(x^2+1\right)+\frac{2}{x^2+1}\ge2\sqrt{\left(x^2+1\right).\frac{2}{x^2+1}}=2\sqrt{2}\)
Vậy \(P\ge2\sqrt{2}+1\)
Dấu bằng xảy ra khi
\(x^2+1=\frac{2}{x^2+1}\Leftrightarrow x^2+1=\sqrt{2}\Rightarrow x^2=\sqrt{2}-1\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\sqrt{2}-1}\\x=-\sqrt{\sqrt{2}-1}\end{cases}}\)
