a: x1+x2=-2; x1x2=-4

x1+x2+2+2=-2+2+2=2

(x1+2)(x2+2)=x1x2+2(x1+x2)+4

=-4+2*(-2)+4=-4

Phương trình cần tìm là x^2-2x-4=0

b: \(\dfrac{1}{x_1+1}+\dfrac{1}{x_2+1}=\dfrac{x_1+x_2+2}{\left(x_1+1\right)\left(x_2+1\right)}\)

\(=\dfrac{x_1+x_2+2}{x_1x_2+\left(x_1+x_2\right)+1}\)

\(=\dfrac{-2+2}{-4+\left(-2\right)+1}=0\)

\(\dfrac{1}{x_1+1}\cdot\dfrac{1}{x_2+1}=\dfrac{1}{x_1x_2+x_1+x_2+1}=\dfrac{1}{-4-2+1}=\dfrac{-1}{5}\)

Phương trình cần tìm sẽ là; x^2-1/5=0

c: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=\dfrac{x_1^2+x_2^2}{x_1x_2}=\dfrac{\left(-2\right)^2-2\cdot\left(-4\right)}{-4}=\dfrac{4+8}{-4}=-3\)

x1/x2*x2/x1=1

Phương trình cần tìm sẽ là:

x^2+3x+1=0

\(ac=-3< 0\Rightarrow\) pt đã cho luôn có 2 nghiệm pb trái dấu với mọi m

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

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

\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}{9}=m-1\)

\(\Leftrightarrow8\left(m-1\right)^3+18\left(m-1\right)=9\left(m-1\right)\)

\(\Leftrightarrow\left(m-1\right)\left[8\left(m-1\right)^2+9\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\8\left(m-1\right)^2+9=0\left(vô-nghiệm\right)\end{matrix}\right.\)

a) Thay m=-2 vào phương trình, ta được:

\(x^2-\left(-x\right)-2=0\)

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

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

Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=1;x_2=\dfrac{c}{a}=\dfrac{-2}{1}=-2\)

\(x^2-4x-6=0\)

\(\text{Δ}=\left(-4\right)^2-4\cdot1\cdot\left(-6\right)=16+24=40>0\)

=>Phương trình này có hai nghiệm phân biệt

Theo vi-et, ta có:

\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4;x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-6}{1}=-6\)

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

\(=4^2-2\cdot\left(-6\right)=16+12=28\)

\(B=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1\cdot x_2}=\dfrac{4}{-6}=-\dfrac{2}{3}\)

\(C=x_1^3+x_2^3\)

\(=\left(x_1+x_2\right)^3-3\cdot x_1\cdot x_2\cdot\left(x_1+x_2\right)\)

\(=4^3-3\cdot4\cdot\left(-6\right)=64+72=136\)

\(D=\left|x_1-x_2\right|\)

\(=\sqrt{\left(x_1-x_2\right)^2}\)

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

\(=\sqrt{4^2-4\cdot\left(-6\right)}=\sqrt{16+24}=\sqrt{40}=2\sqrt{10}\)

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

Thay \(m=3\) vào \(\left(1\right)\)

\(\Rightarrow x^2+3x+3-1=0\)

\(\Rightarrow x^2+3x+2=0\)

\(\Rightarrow x^2+x+2x+2=0\)

\(\Rightarrow x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)

Vậy \(S=\left\{-2;-1\right\}\) khi \(m=3\)

Bài giải 

Ta có : \(\hept{\begin{cases}x_1.x_2=m^2+3\\x_1+x_2=2\left(m+1\right)\end{cases}}\)

\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{8}{x_1.x_2}\Leftrightarrow\frac{x_1^2+x_2^2}{x_1.x_2}=\frac{8}{x_1.x_2}\)

<=> ( x₁ + x₂ ) ² -2x₁x₂ = 8

<=>4(m+1)² -2(m²+ 3 ) = 8 <=> 2m² + 8m - 10=0

<=> \(\orbr{\begin{cases}m=1\\m=-5\left(L\right)\end{cases}}\)