\(\text{Δ}=2^2-4\cdot1\cdot m=4-4m\)

Để phương trình có hai nghiệm thì Δ>=0

=>-4m+4>=0

=>-4m>=-4

=>m<=1(1)

Theo Vi-et, ta có: 

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

\(\dfrac{x_1^2-3x_1+m}{x_2}+\dfrac{x_2^2-3x_2+m}{x_1}< =2\)

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

=>\(\dfrac{\left(x_1+x_2\right)^3-3x_1x_2-3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+m\left(x_1+x_2\right)}{x_1x_2}< =2\)

=>\(\dfrac{\left(-2\right)^3-3\cdot m-3\left[\left(-2\right)^2-2m\right]+m\cdot\left(-2\right)}{m}< =2\)

=>\(\dfrac{-8-3m-3\left(4-2m\right)-2m}{m}-2< =0\)

=>\(\dfrac{-5m-8-12+6m}{m}-2< =0\)

=>\(\dfrac{m-20-2m}{m}< =0\)

=>\(\dfrac{-m-20}{m}< =0\)

=>\(\dfrac{m+20}{m}>=0\)

=>\(\left[{}\begin{matrix}m>0\\m< =-20\end{matrix}\right.\)

Kết hợp (1), ta được: \(\left[{}\begin{matrix}0< m< =1\\m< =-20\end{matrix}\right.\)

a, \(\Delta=m^2-4\left(-4\right)=m^2+16\)> 0 

Vậy pt luôn có 2 nghiệm pb 

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

Ta có \(\left(x_1+x_2\right)^2-2x_1x_2=5\)

Thay vào ta được \(m^2-2\left(-4\right)=5\Leftrightarrow m^2+3=0\left(voli\right)\)

Đăng lại lớp đi chụy :)

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

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-1\end{matrix}\right.\)

\(x^4_1-x^3_1=x^4_2-x^3_2\\ \Leftrightarrow\left(x^4_1-x_2^4\right)-\left(x^3_1+x^3_2\right)=0\\ \Leftrightarrow\left(x^2_1-x^2_2\right)\left(x^2_1+x^2_2\right)-\left(x_1+x_2\right)\left(x^2_1+x^2_2-x_1x_2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=0\\ \Leftrightarrow\left(m-1\right).2\left[2^2-2\left(m-1\right)\right]-2\left[2^2-3\left(m-1\right)\right]=0\)

\(\Leftrightarrow\left(2m-2\right)\left(4-2m+2\right)-2\left(4-3m+3\right)=0\)

\(\Leftrightarrow\left(2m-2\right)\left(6-2m\right)-2\left(7-3m\right)=0\)

\(\Leftrightarrow...\)

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

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-1\end{matrix}\right.\)

\(x^4_1-x^3_1=x^4_2-x^3_2\\ \Leftrightarrow\left(x^4_1-x_2^4\right)-\left(x^3_1-x^3_2\right)=0\\ \Leftrightarrow\left(x^2_1-x^2_2\right)\left(x^2_1+x^2_2\right)-\left(x_1-x_2\right)\left(x^2_1+x^2_2+x_1x_2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]=0\)

\(\Leftrightarrow\left(x_1-x_2\right).2\left(4-2m+2\right)-\left(x_1-x_2\right)\left(4-m+1\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right).2\left(6-2m\right)-\left(x_1-x_2\right)\left(5-m\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(12-4m-5+m\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(7-3m\right)=0\)

\(\Leftrightarrow...\)

Để pt có 2 nghiệm phân biệt thì \(\Delta'=m^2-\left(m+2\right)>0\Leftrightarrow\left(m+1\right)\left(m-2\right)>0\Leftrightarrow\left[{}\begin{matrix}m>2\\m< -1\end{matrix}\right.\). (1)

Khi đó theo hệ thức Viète ta có \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m+2\end{matrix}\right.\).

Ta có \(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\left(2m\right)^3-3.2m.\left(m+2\right)=8m^3-6m^2-12m\).

Do đó \(8m^3-6m^2-12m\le16\Leftrightarrow\left(m-2\right)\left(8m^2+10m+8\right)\le0\Leftrightarrow m\le2\)

(do \(8m^2+10m+8=2\left(2m+\dfrac{5}{4}\right)^2+\dfrac{39}{8}>0\forall m\)).

Kết hợp vs (1) ta có m < -1.

a, x = 3 , x= -1

b, m = 3 , m = 1

a: Thay m=-5 vào (1), ta được:

\(x^2+2\left(-5+1\right)x-5-4=0\)

\(\Leftrightarrow x^2-8x-9=0\)

=>(x-9)(x+1)=0

=>x=9 hoặc x=-1

b: \(\text{Δ}=\left(2m+2\right)^2-4\left(m-4\right)=4m^2+8m+4-4m+16=4m^2+4m+20>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt 

\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-3\)

\(\Leftrightarrow x_1^2+x_2^2=-3x_1x_2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow\left(2m+2\right)^2+m-4=0\)

\(\Leftrightarrow4m^2+9m=0\)

=>m(4m+9)=0

=>m=0 hoặc m=-9/4

a, với =-3

\(=>x^2-6x+6=0\)

\(\Delta=\left(-6\right)^2-4.6=12>0\)

=>pt có 2 nghiệm phân biệt x3,x4

\(=>\left[{}\begin{matrix}x3=\dfrac{6+\sqrt{12}}{2}=3+\sqrt{3}\\x4=\dfrac{6-\sqrt{12}}{2}=3-\sqrt{3}\end{matrix}\right.\)

b, \(\Delta=\left(2m\right)^2-4\left(m^2+m\right)=4m^2-4m^2-4m=-4m\)

pt đã cho đề bài có 2 nghiệm phân biệt x1,x2 khi

\(-4m>0< =>m< 0\)

theo vi ét \(=>\left\{{}\begin{matrix}x1+x2=-2m\\x1x2=m^2+m\end{matrix}\right.\)

có \(\left(x1-x2\right)\left(x1^2-x2^2\right)=32\)

\(< =>\left(x1-x2\right)^2\left(x1+x2\right)=32\)

\(< =>\left[x1^2-2x1x2+x2^2\right]\left(-2m\right)=32\)

\(< =>\left[\left(x1+x2\right)^2-4x1x2\right]\left(-2m\right)=32\)

\(< =>\left[\left(-2m\right)^2-4\left(m^2+m\right)\right]\left(-2m\right)=32< =>m=2\)(loại)

Vậy \(m\in\varnothing\)

Lời giải:
a. Với $m=-3$ thì pt trở thành:

$x^2-6x+6=0\Leftrightarrow x=3\pm \sqrt{3}$

b. Để pt có 2 nghiệm thì: $\Delta'=m^2-(m^2+m)=-m\geq 0$

$\Leftrightarrow m\leq 0$

Áp dụng định lý Viet: $x_1+x_2=-2m; x_1x_2=m^2+m$

Khi đó:
$(x_1-x_2)(x_1^2-x_2^2)=32$

$\Leftrightarrow (x_1-x_2)^2(x_1+x_2)=32$

$\Leftrightarrow [(x_1+x_2)^2-4x_1x_2](x_1+x_2)=32$

$\Leftrightarrow [(-2m)^2-4(m^2+m)](-2m)=32$

$\Leftrightarrow 8m^2=32$

$\Leftrightarrow m^2=4$

$\Rightarrow m=-2$ (do $m\leq 0$)

Vây.........

a) Thay m=-3 vào phương trình, ta được:

\(x^2-6x+\left(-3\right)^2+\left(-3\right)=0\)

\(\Leftrightarrow x^2-6x+6=0\)

\(\Delta=\left(-6\right)^2-4\cdot6=36-24=12\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=\dfrac{6+2\sqrt{3}}{2}=3+\sqrt{3}\end{matrix}\right.\)

\(\Delta'=\left(m-1\right)^2+m^3-\left(m+1\right)^2=m^3-4m\ge0\) \(\Rightarrow\left[{}\begin{matrix}m\ge2\\-2\le m\le0\end{matrix}\right.\)

Theo hệ thức Viet:  \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m^3+\left(m+1\right)^2\end{matrix}\right.\)

Do \(x_1+x_2\le4\Rightarrow m-1\le2\Rightarrow m\le3\)

\(\Rightarrow\left[{}\begin{matrix}2\le m\le3\\-2\le m\le0\end{matrix}\right.\)

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

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

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

\(=8\left(5m-2m^2\right)\)

\(P=8\left(5m-2m^2-2+2\right)=16-8\left(m-2\right)\left(2m-1\right)\le16\)

\(P_{max}=16\) khi \(m=2\)

\(P=8\left(5m-2m^2+18-18\right)=8\left(9-2m\right)\left(m+2\right)-144\ge-144\)

\(P_{min}=-144\) khi \(m=-2\)

\(x^2-2\left(m+4\right)x+m^2+8m-9=0\left(1\right)\)

Ta giải \(\Delta=[-2\left(m+4\right)]^2-4\left(m^2+8m-9\right)=100>0\forall m\)

suy ra pt có 2 nghiệm phân biệt \(x_1,x_2\forall m\).

Ta có: \(x_1=m-1\), \(x_2=m+1\) (thay \(\Delta\) vào công thức tìm nghiệm phân biệt).

Gọi \(A=\dfrac{x_1^2+x_2^2-48}{x_1^2+x_2^2}\).

\(\Rightarrow A=1-\dfrac{48}{x_1^2+x_2^2}=1-\dfrac{48}{\left(m-1\right)^2+\left(m+1\right)^2}=1-\dfrac{24}{m^2+1}\).

Để biểu thức A nguyên thì \(\dfrac{24}{m^2+1}\) nguyên, suy ra \(m^2+1\inƯ\left(24\right)\).

\(\Rightarrow m^2+1\in\left\{1;2;4;6;8;12;24\right\}\)

\(\Rightarrow m\in\left\{0;\pm1\right\}\) (vì m nhận giá trị nguyên)

Vậy \(m\in\left\{0;\pm1\right\}\) là giá trị cần tìm.

Mình chỉnh sửa lại một chút nhé.

\(A=1-\dfrac{24}{m^2+2}\)

\(\Rightarrow...\)\(\Rightarrow\)\(m^2+2\in\left\{1;2;3;4;6;8;12;24\right\}\)

\(\Rightarrow m\in\left\{0;\pm1;\pm2\right\}\)

Vậy...