P/t có : \(\Delta\)' = \(\left(m-1\right)^2-\left(-2m\right)=m^2+1\ge1>0\forall m\)  -> P/t có 2 no x1 ; x2 p/b . Theo Viet có : \(\left\{{}\begin{matrix}x1+x2=2\left(m-1\right)\\x1.x2=-2m\end{matrix}\right.\) 

\(\Rightarrow x1+x2+x1.x2=-2\) 

Mặt # ta có : \(\left[{}\begin{matrix}x1=m-1+\sqrt{m^2+1};x2=m-1-\sqrt{m^2+1}\\x1=m-1-\sqrt{m^2+1};x2=m-1+\sqrt{m^2+1}\end{matrix}\right.\)

Ta có : \(x1^2+3x2-4x1.x2=5\) 

Đặt x1 = a ; x2 = b ; ta có hệ : \(\left\{{}\begin{matrix}a+b+ab+2=0\left(1\right)\\a^2+3b-4ab-5=0\left(2\right)\end{matrix}\right.\)

Từ (1) suy ra : \(b=\dfrac{-\left(a+2\right)}{a+1}\)  ; ab = -a-b-2  ( Loại a = -1)

Thay vào (2) được : \(a^2+3b+4a+4b+8-5=0\)  \(\Leftrightarrow\left(a+2\right)^2+7b=1\)

\(\Leftrightarrow b=\dfrac{1-\left(a+2\right)^2}{7}\)

Suy ra : \(\dfrac{-\left(a+2\right)}{a+1}=\dfrac{1-\left(a+2\right)^2}{7}\)

\(\Leftrightarrow7\left(a+2\right)=\left[\left(a+2\right)^2-1\right]\left(a+1\right)\)

\(\Leftrightarrow\) \(a^3+5a^2-11=0\)

Đoạn này bí quá ; bn thử giải xem 

Moij người giúp mình với ạ mình đang cần gấp ạ

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-4\right)\\x_1x_2=-m^2+4\end{matrix}\right.\)

\(\dfrac{x_1+x_2}{x_1x_2}+\dfrac{4}{x_1x_2}=1\)

Thay vào ta được : \(\dfrac{2\left(m-4\right)+4}{-m^2+4}=1\Leftrightarrow\dfrac{2m-4}{\left(2-m\right)\left(m+2\right)}=1\Leftrightarrow\dfrac{-2}{m+2}=1\Rightarrow-2=m+2\Leftrightarrow m=-4\)

1.

\(a+b+c=0\) nên pt luôn có 2 nghiệm

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

\(A=\dfrac{2x_1x_2+3}{x_1^2+x_2^2+2x_1x_2+2}=\dfrac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\dfrac{2\left(m-1\right)+3}{m^2+2}=\dfrac{2m+1}{m^2+2}\)

\(A=\dfrac{m^2+2-\left(m^2-2m+1\right)}{m^2+2}=1-\dfrac{\left(m-1\right)^2}{m^2+2}\le1\)

Dấu "=" xảy ra khi \(m=1\)

2.

\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\) nên pt luôn có 2 nghiệm pb

Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)

\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Rightarrow\dfrac{\left(m-2\right)^2-2m^2+4\left(m-2\right)+4}{m-2-m+1}=4\)

\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)

Ta có: \(x_1^2+x_2^2+3x_1x_2=2\)

=>\(x_1^2+x_2^2+2x_1x_2+x_1x_2=2\)

=>\(\left(x_1+x_2\right)^2+x_1x_2=2\)

a: \(\Delta=\left(-4\right)^2-4\cdot1\cdot\left(m-1\right)\)

=16-4m+4

=-4m+20

Để phương trình có hai nghiệm thì -4m+20>=0

=>-4m>=-20

=>m<=5

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=4;x_1x_2=\frac{c}{a}=m-1\)

\(\left(x_1+x_2\right)^2+x_1x_2=2\)

=>\(4^2+\left(m-1\right)=2\)

=>m-1+16=2

=>m+15=2

=>m=-13(nhận)

b: \(\Delta=\left(-2\right)^2-4\cdot1\cdot\left(m-1\right)\)

=4-4m+4

=-4m+8

Để phương trình có hai nghiệm thì -4m+8>=0

=>-4m>=-8

=>m<=2

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2;x_1x_2=\frac{c}{a}=m-1\)

\(\left(x_1+x_2\right)^2+x_1x_2=2\)

=>\(2^2+m-1=2\)

=>m+3=2

=>m=-1(nhận)

b, Ta có : \(0\le x\le1\)

\(\Rightarrow-2\le x-2\le-1< 0\)

Ta có : \(y=f\left(x\right)=2\left(m-1\right)x+\dfrac{m\left(x-2\right)}{\left(2-x\right)}\)

\(=2\left(m-1\right)x-m< 0\)

TH1 : \(m=1\) \(\Leftrightarrow m>0\)

TH2 : \(m\ne1\) \(\Leftrightarrow x< \dfrac{m}{2\left(m-1\right)}\)

Mà \(0\le x\le1\)

\(\Rightarrow\dfrac{m}{2\left(m-1\right)}>1\)

\(\Leftrightarrow\dfrac{m-2\left(m-1\right)}{2\left(m-1\right)}>0\)

\(\Leftrightarrow\dfrac{2-m}{m-1}>0\)

\(\Leftrightarrow1< m< 2\)

Kết hợp TH1 => m > 0

Vậy ...
 

\(x^2-2\left(m-1\right)x-m^3+\left(m+1\right)^2=0\)

Để pt có hai nghiệm thỏa mãn

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\x_1+x_2=2\left(m-1\right)\le4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m-2\right)\left(m+2\right)\ge0\\m\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m\in\left[-2;0\right]\cup\left(2;+\infty\right)\cup\left\{2\right\}\\m\le3\end{matrix}\right.\)\(\Rightarrow m\in\left[-2;0\right]\cup\left[2;3\right]\)

\(P=x^3_1+x_2^3+x_1x_2\left(3x_1+3x_2+8\right)\)

\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_1\left(x_1+x_2\right)+8x_1x_2\)

\(=8\left(m-1\right)^3+8\left(-m^3+m^2+2m+1\right)\)

\(=-16m^2+40m\)

Vẽ BBT với \(f\left(m\right)=-16m^2+40m\) ;\(m\in\left[-2;0\right]\cup\left[2;3\right]\)

Tìm được \(f\left(m\right)_{min}=-144\Leftrightarrow m=-2\)

\(f\left(m\right)_{max}=16\Leftrightarrow m=2\)

\(\Rightarrow P_{max}=16;P_{min}=-144\)

Vậy....