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

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

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

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

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

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

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

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

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

=>Phương trình này có hai nghiệm phân biệt

Theo vi-et, ta có:

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

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

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

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

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

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

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

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

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

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

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

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

\(x^2-2nx+3n+3=\left(x-n\right)^2-\left(n^2-3n+3\right)=0\)\(\left(x-n\right)^2=\left(n-\frac{3}{2}\right)^2+\frac{3}{4}=\frac{\left(2n-3\right)^2+3}{4}>0\forall n\) vậy luôn tồn tại hai nghiệm

\(\orbr{\begin{cases}x_1=\frac{n-\sqrt{\left(2n-3\right)^2+3}}{2}\\x_2=\frac{n+\sqrt{\left(2n-3\right)^2+3}}{2}\end{cases}}\)

a) \(\frac{x_1}{x_2}=\frac{4x_1-x_2}{x_1}\Leftrightarrow\frac{x_1^2-4x_1x_2+x_2^2}{x_1x_2}=0\)

\(x_1x_2=n^2-\frac{\left(2n-3\right)^2+3}{4}=\frac{4n^2-4n^2+12n-9-3}{4}=3n-3\)

với n=1 hay m=0 : Biểu thức cần C/m không tồn tại => xem lại đề

a: \(\text{Δ}=\left[-\left(m+3\right)\right]^2-4\cdot2\cdot m\)

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

\(=m^2-2m+9=\left(m-1\right)^2+8>0\forall m\)

=>Phương trình (1) luôn có hai nghiệm phân biệt

b: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{m+3}{2}\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m}{2}\end{matrix}\right.\)

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

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

\(=\sqrt{\dfrac{1}{4}\left(m+3\right)^2-4\cdot\dfrac{m}{2}}\)

\(=\sqrt{\dfrac{1}{4}\left(m^2+6m+9\right)-2m}\)

\(=\sqrt{\dfrac{1}{4}m^2+\dfrac{3}{2}m+\dfrac{9}{4}-2m}\)

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

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

\(=\sqrt{\dfrac{1}{4}\left(m^2-2m+1+8\right)}\)

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

Dấu '=' xảy ra khi m-1=0

=>m=1

a)Có ac=-1<0

=>pt luôn có hai nghiệm trái dấu

b)Do x₁;x₂ là hai nghiệm của pt

=> \(\left\{{}\begin{matrix}x_1^2-mx_1-1=0\\x_2^2-mx_2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1^2-1=mx_1\\x_2^2-1=mx_2\end{matrix}\right.\)

=>\(P=\dfrac{mx_1+x_1}{x_1}-\dfrac{mx_2+x_2}{x_2}\)\(=m+1-\left(m+1\right)=0\)

\(2x^2+3mx-\sqrt{2}=0\)

Phương trình có 2 nghiệm phân biệt <=> \(\Delta=\left(3m\right)^2-4\cdot2\cdot\left(\sqrt{2}\right)>0\)

<=> \(9m^2+3\sqrt{2}>0\)(luôn đúng)

=> PT có 2 nghiệm phân biệt x₁;x₂ với mọi m \(\hept{\begin{cases}x_1+x_2=\frac{-3m}{2}\\x_1x_2=\frac{-\sqrt{2}}{2}\end{cases}}\)

\(M=\left(x_1-x_2\right)^2+\left(\frac{1+x_1^2}{x_1}-\frac{1+x_2^2}{x_2}\right)\)

\(=x_1^2+x_2^2-2x_1x_2+\left[\frac{x_2\left(1+x_1^2\right)-x_1\left(1+x_2^2\right)}{x_1x_2}\right]^2\)

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

\(=\left(\frac{-3m}{2}\right)^2-4\cdot\left(\frac{\sqrt{2}}{2}\right)+\frac{\left(x_2-x_1\right)^2\cdot\left(1+x_1x_2\right)^2}{\left(x_1x_2\right)^2}\)

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

\(=\frac{9m^2}{4}+2\sqrt{2}+\left(\frac{9m^2}{4}+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)\)

\(=\frac{9m^2}{4}\left(4-2\sqrt{2}\right)+2\sqrt{2}\left(4-2\sqrt{2}\right)\ge2\sqrt{2}\left(4-2\sqrt{2}\right)\ge8\sqrt{2}-8\)

Dấu "=" xảy ra <=> m=0

Em xem lại dòng thứ 3 sau khi M = nhé Linh !

Dòng thứ 3 phải sửa là:

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

Đúng không ạ?!

\(\Delta=4m^2-4m+1-4\left(2m-2\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0\)

Do đó pt luôn có nghiệm

Theo định lí Vi-ét:

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

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

\(A=\left(2m-1\right)^2-2\left(2m-2\right)\)           

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

\(A=4m^2-8m+5\)

\(A=4\left(m-1\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\) m=1

Tick hộ nha 😘

pt có nghiệm \(< =>\Delta\ge0\)

\(< =>[-\left(2m-1\right)]^2-4\left(2m-2\right)\ge0\)

\(< =>4m^2-4m+1-8m+8\ge0\)

\(< =>4m^2-12m+9\ge0\)

\(< =>4\left(m^2-3m+\dfrac{9}{4}\right)\ge0\)

\(=>m^2-2.\dfrac{3}{2}m+\dfrac{9}{4}\ge0< =>\left(m-\dfrac{2}{3}\right)^2\ge0\)(luôn đúng)

=>pt luôn có 2 nghiệm 

theo vi ét \(=>\left\{{}\begin{matrix}x1+x2=2m-1\\x1x2=2m-2\end{matrix}\right.\)

\(A=\left(x1+x2\right)^2-2x1x2=\left(2m-1\right)^2-2\left(2m-2\right)\)

\(A=4m^2-4m+1-4m+4=4m^2+5\ge5\)

dấu"=" xảy ra<=>m=0

\(\Delta=\left(2m-1\right)^2-4\left(2m-2\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0;\forall m\)

\(\Rightarrow\) Phương trình đã cho luôn có nghiệm

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

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

\(A=\left(2m-1\right)^2-2\left(2m-2\right)\)

\(A=4m^2-8m+5=4\left(m-1\right)^2+1\ge1\)

Dấu "=" xảy ra khi \(m-1=0\Leftrightarrow m=1\)