\(\text{Δ}=\left[-\left(m+1\right)\right]^2-4\cdot1\cdot m\)

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

\(=\left(m-1\right)^2>=0\forall m\)

=>Phương trình luôn có hai nghiệm

Theo Vi-et, ta có:

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

\(x_1^2+x_2^2=\left(x_1-1\right)\left(x_2-1\right)-x_1-x_2+5\)

=>\(\left(x_1+x_2\right)^2-2x_1x_2=x_1x_2-2\left(x_1+x_2\right)+6\)

=>\(\left(m+1\right)^2-2m=m-2\left(m+1\right)+6\)

=>\(m^2+1=m-2m-2+6\)

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

=>\(m^2+m-3=0\)

=>\(m=\dfrac{-1\pm\sqrt{13}}{2}\)

Moij người giúp mình với ạ mình đang cần gấp ạ

​\(\Delta'=m^2+1\Rightarrow\left\{{}\begin{matrix}x_1=m+1+\sqrt{m^2+1}\\x_2=m+1-\sqrt{m^2+1}\end{matrix}\right.\)

(Do \(m+1-\sqrt{m^2+1}< \sqrt{m^2+1}+1-\sqrt{m^2+1}< 4\) nên nó ko thể là nghiệm \(x_1\))

Từ điều kiện \(x_1\ge4\Rightarrow m+1+\sqrt{m^2+1}\ge4\Rightarrow\sqrt{m^2+1}\ge3-m\)

\(\Rightarrow\left[{}\begin{matrix}m\ge3\\\left\{{}\begin{matrix}m< 3\\m^2+1\ge m^2-6m+9\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m\ge\dfrac{4}{3}\)

​​Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{matrix}\right.\)

\(x_1^2=9x_2+10\Leftrightarrow x_1\left(x_1+x_2\right)-x_1x_2=9x_2+10\)

\(\Leftrightarrow2\left(m+1\right)x_1-2m=9x_2+10\)

\(\Leftrightarrow2\left(m+1\right)x_1-2m=9\left(2\left(m+1\right)-x_1\right)+10\)

\(\Leftrightarrow\left(2m+11\right)x_1=20m+28\Rightarrow x_1=\dfrac{20m+28}{2m+11}\) 

\(\Rightarrow x_2=2\left(m+1\right)-x_1=\dfrac{4m^2+6m-6}{2m+11}\)

Thế vào \(x_1x_2=2m\)

\(\Rightarrow\left(\dfrac{20m+28}{2m+11}\right)\left(\dfrac{4m^2+6m-6}{2m+11}\right)=2m\)

\(\Leftrightarrow\left(3m-4\right)\left(12m^2+40m+21\right)=0\)

\(\Leftrightarrow m=\dfrac{4}{3}\) (do \(12m^2+40m+21>0;\forall m\ge\dfrac{4}{3}\))

|x1|=3|x2|

=>|2m+2-x2|=|3x2|

=>4x2=2m+2 hoặc -2x2=2m+2

=>x2=1/2m+1/2 hoặc x2=-m-1

Th1: x2=1/2m+1/2

=>x1=2m+2-1/2m-1/2=3/2m+3/2

x1*x2=m^2+2m

=>1/2(m+1)*3/2(m+1)=m^2+2m

=>3/4m^2+3/2m+3/4-m^2-2m=0

=>m=1 hoặc m=-3

TH2: x2=-m-1 và x1=2m+2+m+1=3m+3

x1x2=m^2+2m

=>-3m^2-6m-3-m^2-2m=0

=>m=-1/2; m=-3/2

Vì \(a\cdot c=1\cdot\left(-2\right)=-2< 0\)

nên phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m\\x_1x_2=\dfrac{c}{a}=-2\end{matrix}\right.\)

Sửa đề: \(x_1^2\cdot x_2+x_1\cdot x_2^2+7>x_1^2+x_2^2+\left(x_1+x_2\right)^2\)

=>\(x_1x_2\left(x_1+x_2\right)+7>\left(x_1+x_2\right)^2-2x_1x_2+\left(x_1+x_2\right)^2\)

=>\(-2m+7>m^2-2\left(-2\right)+m^2\)

=>\(2m^2+4< -2m+7\)

=>\(2m^2+2m-3< 0\)

=>\(\dfrac{-1-\sqrt{7}}{2}< m< \dfrac{-1+\sqrt{7}}{2}\)

Cái này phân tích đề ra là lm được bạn nhé

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 

\(\Delta=1-4m>0\Rightarrow m< \dfrac{1}{4}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=m\end{matrix}\right.\)

\(\left(x_1^2+x_2+m\right)\left(x_2^2+x_1+m\right)=m^2-m-1\)

\(\Leftrightarrow\left[x_1\left(x_1+x_2\right)-x_1x_2+x_2+m\right]\left[x_2\left(x_1+x_2\right)-x_1x_2+x_1+m\right]=m^2-m-1\)

\(\Leftrightarrow\left(x_1+x_2\right)\left(x_1+x_2\right)=m^2-m-1\)

\(\Leftrightarrow m^2-m-1=1\)

\(\Leftrightarrow m^2-m-2=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=2>\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)