\(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot\left(m+2\right)\)

\(=25-4m-8=-4m+17\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>-4m+17>0

=>-4m>-17

=>\(m< \dfrac{17}{4}\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-5\right)}{1}=5\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m+2}{1}=m+2\end{matrix}\right.\)

\(P=x_1^2\cdot x_2+x_1\cdot x_2^2-x_1^2\cdot x_2^2-4\)

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

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

\(=5m+10-m^2-4m-4-4\)

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

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

\(=-\left(m^2-m+\dfrac{1}{4}-\dfrac{9}{4}\right)\)

\(=-\left(m-\dfrac{1}{2}\right)^2+\dfrac{9}{4}< =\dfrac{9}{4}\forall m\)

Dấu '=' xảy ra khi \(m=\dfrac{1}{2}\)

\(\Delta=25-4\left(m+2\right)=17-4m>0\Rightarrow m< \dfrac{17}{4}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m+2\end{matrix}\right.\)

\(P=x_1x_2\left(x_1+x_2\right)-\left(x_1x_2\right)^2-4\)

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

\(=-\left[\left(m+2\right)-\dfrac{5}{2}\right]^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(P_{max}=\dfrac{9}{4}\) khi \(m+2=\dfrac{5}{2}\Rightarrow m=\dfrac{1}{2}\)

Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{1}{1}=1\\x_1x_2=\dfrac{c}{a}=-\dfrac{3}{1}=-3\end{matrix}\right.\)

a

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

\(=\left(x_1+x_2\right)^2-2x_1x_2=1^2-2.\left(-3\right)=1+6=7\)

b

\(B=x_1^2x_2+x_1x_2^2=x_1x_2\left(x_1+x_2\right)=\left(-3\right).1=-3\)

c

\(C=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{-3}=-\dfrac{1}{3}\)

d

\(D=\dfrac{x_2}{x_1}+\dfrac{x_1}{x_2}=\dfrac{x_2^2}{x_1x_2}+\dfrac{x_1^2}{x_1x_2}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{1^2-2.\left(-3\right)}{-3}=\dfrac{1+6}{-3}=\dfrac{7}{-3}=-\dfrac{3}{7}\)

\(\Delta'=\left(2m+1\right)^2-\left(4m^2+4m\right)=1>0;\forall m\Rightarrow\) pt luôn có 2 nghiệm pb

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(2m+1\right)\\x_1x_2=4m^2+4m\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2\ge0\\\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(2m+1\right)\ge0\\-2x_1x_2=2x_1x_2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ge-\dfrac{1}{2}\\x_1x_2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ge-\dfrac{1}{2}\\4m^2+4m=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m=0\\mm=-1< -\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)

áp dụng vi et 
x1+x2=\(\dfrac{-b}{a}=4m+2\)
x1.x2=\(\dfrac{c}{a}=4m^2+4m\)
ta có :
\(|x_1-x_2|=x_1+x_2\)
<->(x1-x2)²=(x1+x2)²
<->(x1+x2)²-4x₁.x₂=(4m+2)²
<->(4m+2)²-4(4m²+4m)=(4m+2)²
<->16m²+4+16m-16m²-16m=16m²+4+16m
<->16m²+16m=0
<->16m(m+1)=0
<->m=0
     m=-1
vậy m =0 và m=-1 thì tm hệ thức trên

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

\(\Rightarrow T=x_1+x_2+3x_1x_2=-2+3.\left(-1\right)=-5\)

Theo Vi-et, ta có: 

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

Ta có: \(T=x_1+x_2+3x_1x_2\)

\(=-2+3\cdot\left(-1\right)\)

=-5

\(\text{Δ}=\left(-m\right)^2-4\left(m-5\right)\)

\(=m^2-4m+20\)

\(=m^2-4m+4+16=\left(m-2\right)^2+16>0\forall m\)

=>Phương trình luôn có 2 nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-m\right)}{1}=m\\x_1\cdot x_2=\dfrac{c}{a}=m-5\end{matrix}\right.\)

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

\(x_1\cdot x_2=m-5\)

=>\(\left(1-m\right)\left(2m-1\right)=m-5\)

=>\(2m-1-2m^2+m-m+5=0\)

=>\(-2m^2+2m+4=0\)

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

=>(m-2)(m+1)=0

=>\(\left[{}\begin{matrix}m-2=0\\m+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\left(nhận\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)

Nếu đề bài là

Tính P=\(\frac{x_1^2+x_1-1}{x_1}\)-\(\frac{x_2^2+x_2-1}{x_2}\)

Thì lời giải như sau:

Theo định lý Viete, ta có:

x₁.x₂=-1

Khi đó P=\(\frac{x_1^2+x_1+x_1.x_2}{x_1}\)-\(\frac{x_2^2+x_2+x_1.x_2}{x_2}\)

Do x₁ và x₂ không thể bằng không nên ta chia tử mẫu của mỗi hạng tử cho x₁,x₂

Khi đó P=x₁+x₂+1-(x₂+x₁+1)=0

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{20a-11}{2012}\\x_1x_2=-1\end{matrix}\right.\)

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

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

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

\(=6\left(x_1-x_2\right)^2=6\left(x_1+x_2\right)^2-24x_1x_2\)

\(=6\left(\dfrac{20a-11}{2012}\right)^2+24\ge24\)

Dấu "=" xảy ra khi \(a=\dfrac{11}{20}\)

Áp dụng hệ thức Vi-et, ta được:

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

Ta có: \(\dfrac{1}{x_1}+\dfrac{1}{x_2}\)

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

\(=\dfrac{5}{-1}=-5\)