\(\left(m+1\right)x^2+2mx+m-1=0\)

\(\Delta'=m^2-\left(m-1\right)\left(m+1\right)=m^2-m^2+1=1>0\)

=> Ptr luôn có 2 nghiệm phân biệt

Theo vi-et : \(\hept{\begin{cases}x_1+x_2=\frac{-2m}{m+1}\\x_1.x_2=\frac{m-1}{m+1}\end{cases}}\)

<=> \(\left(x_1+x_2\right)^2-2x_1x_2=\left(\frac{-2m}{m+1}\right)^2-\frac{2\left(m-1\right)}{m+1}\)

<=> \(x_1^2+x_2^2=\frac{4m^2}{\left(m+1\right)^2}-\frac{2\left(m-1\right)}{m+1}\)

Từ đề bài => \(5=\frac{4m^2-2\left(m-1\right)\left(m+1\right)}{\left(m+1\right)^2}=\frac{4m^2-2m^2+2}{\left(m+1\right)^2}=\frac{2m^2+2}{\left(m+1\right)^2}\)

<=> \(5m^2+10m+5=2m^2+2\)

<=> \(3m^2+10m+3=0\)

<=> \(\left(3m+1\right)\left(m+3\right)=0\)

<=> \(\orbr{\begin{cases}m=-\frac{1}{3}\\m=-3\end{cases}}\)

Vậy ....

\(\overrightarrow{AB}=\left(x_B-x_A;y_B-y_A\right)=\left(4-2;-2-\left(-1\right)\right)=\left(2;-1\right)\)

\(\overrightarrow{AC}=\left(x_C-x_A;y_C-y_A\right)=\left(0-2;3-\left(-1\right)\right)=\left(-2;4\right)\)

\(f\left(x\right)=x+\frac{2}{x-1}\)

\(=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\)

Ap dụng bất đẳng thức Cô - si :

\(f\left(x\right)>2.\sqrt{\frac{x-1}{2}.\frac{2}{x-1}}+\frac{1}{2}=\frac{3}{2}\)

Dấu '' = '' xảy ra khi \(\frac{x-1}{2}=\frac{2}{x-1}\)

\(\left(x-1\right)^2=4\)

\(x-1=2\)

\(x=3\)

Vậy GTNN là 3

Ta có: \(f\left(x\right)=x+\frac{2}{x-1}\) \(=x-1+\frac{2}{x-1}\)\(+1>2\sqrt{\left(x-1\right)\frac{2}{x-1}}+1=2\sqrt{2+1}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x>1\\x-1=\frac{2}{x-1}\end{cases}}\)\(x=1+\sqrt{2}\) vậy \(m=2\sqrt{2}+1\)

no no on onnonoonon

? mình không hiểu nhé