a=1; b=-2m-2; c=-4

Vì ac<0

nên phương trình luôn có hai nghiệm phân biệt

Ptr có: `\Delta = b^2 - 4ac = [-(2m + 1)]^2 - 4 . (-4)`

                                         `= ( 2m + 1)^2 + 16 > 0 AA m`

    `=> \Delta > 0 AA m`

Vật ptr luôn có `2` nghiệm `x_1 , x_2` với mọi `m`

cái này là công thức |a-b|=căn (a-b)^2=căn (a+b)^2-4ab thôi nha bạn

\(\Delta'=\left(m+1\right)^2-\left(5m+1\right)=m^2-3m\ge0\Rightarrow\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=5m+1\end{matrix}\right.\)

a.

\(S=\left(x_1+x_2\right)^2-3x_1x_2=4\left(m+1\right)^2-3\left(5m+1\right)\)

\(=4m^2-7m+1=\dfrac{7}{3}\left(m^2-3m\right)+\dfrac{5}{3}m^2+1\ge1\)

\(S_{min}=1\) khi \(\dfrac{7}{3}\left(m^2-3m\right)+\dfrac{5}{3}m^2=0\Rightarrow m=0\)

b.

\(1< x_1< x_2\Rightarrow\left\{{}\begin{matrix}\left(x_1-1\right)\left(x_2-1\right)>0\\\dfrac{x_1+x_2}{2}>1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1x_2-\left(x_1+x_2\right)+1>0\\x_1+x_2>2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5m+1-2\left(m+1\right)+1>0\\2\left(m+1\right)>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m>0\\m>-1\end{matrix}\right.\) \(\Rightarrow m>0\)

Kết hợp điều kiện delta \(\Rightarrow m\ge3\)

\(a,\Leftrightarrow\Delta\ge0\Leftrightarrow\left(2m+2\right)^2-4\left(5m+1\right)\ge0\Leftrightarrow4m^2-12m\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}m\le0\\m\ge3\end{matrix}\right.\)

\(vi-ét\Rightarrow\left\{{}\begin{matrix}x1+x2=2m+2\\x1x2=5m+1\end{matrix}\right.\)

\(\Rightarrow S=x1^2+x2^2-x1x2=\left(x1+x2\right)^2-3x1x2\)

\(=\left(2m+2\right)^2-3\left(5m+1\right)=4m^2-7m+1\)

\(=\left(2m\right)^2-2.2.\dfrac{7}{4}.m+\left(\dfrac{7}{4}\right)^2-\dfrac{33}{16}=\left(2m-\dfrac{7}{4}\right)^2-\dfrac{33}{16}\left(1\right)\)

\(TH1:m\ge3\Rightarrow\left(1\right)\ge\left(2.3-\dfrac{7}{4}\right)^2-\dfrac{33}{16}=16\)

\(TH2:m\le0\Rightarrow\left(1\right)\ge\left(0-\dfrac{7}{4}\right)^2-\dfrac{33}{16}=1\)

\(\Rightarrow MinS=1\Leftrightarrow m=0\left(tm\right)\)

\(b,1< x1< x2\Leftrightarrow0< x1-1< x2-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\left(x1-1\right)\left(x2-1\right)>0\\x1+x2-2>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x1>1\\x2>1\end{matrix}\right.\\\left\{{}\begin{matrix}x1 < 1\\x2< 1\end{matrix}\right.\end{matrix}\right.\\2m+2-2>0\\\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}x1x2>1\\x1x2< 1\end{matrix}\right.\\m>0\\\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}m>0\\m< 0\end{matrix}\right.\\m>0\\\end{matrix}\right.\Rightarrow m>3\)

Theo Vi-ét \(\hept{\begin{cases}x_1+x_2=-\frac{5}{3}\\x_1x_2=-2\end{cases}}\)

Ta có \(S=y_1+y_2=x_1+x_2+\frac{1}{x_1}+\frac{1}{x_2}=\left(x_1+x_2\right)+\frac{x_1+x_2}{x_1x_2}\)

                                                                           \(=-\frac{5}{3}+\frac{\frac{-5}{3}}{-2}=-\frac{5}{6}\)

       \(P=x_1x_2=\left(x_1+\frac{1}{x_2}\right)\left(x_2+\frac{1}{x_1}\right)=x_1x_2+1+1+\frac{1}{x_1x_2}=-2+2+\frac{1}{-2}=-\frac{1}{2}\)

Khi đó y₁ ; y₂ là nghiệm của pt

\(Y^2-SY+P=0\) 

\(\Leftrightarrow Y^2+\frac{5}{6}Y-\frac{1}{2}=0\)

ko có pt f(x) sao giải đc