\(đk:\left\{{}\begin{matrix}\Delta\ge0\\0< x1\le x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5^2-4\left(-m^2+m+6\right)\ge0\\\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-4m+1=\left(2m-1\right)^2\ge0\left(đúng\right)\\\left\{{}\begin{matrix}5>0đúng\\-m^2+m+6>0\Leftrightarrow-2< m< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-2< m< 3\)

\(\Rightarrow\dfrac{1}{\sqrt{x1}}+\dfrac{1}{\sqrt{x2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{\sqrt{x1}+\sqrt{x2}}{\sqrt{x1x2}}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x1+x2+2\sqrt{x1x2}}{x1x2}=\dfrac{9}{4}\Leftrightarrow\dfrac{5+2\sqrt{-m^2+m+6}}{-m^2+m+6}=\dfrac{9}{4}\)

\(đặt::\sqrt{-m^2+m+6}=t\ge0\Rightarrow\dfrac{5+2t}{t^2}=\dfrac{9}{4}\)

\(\Rightarrow9t^2-8t-20=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{10}{9}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-m^2+m+6}=2\Leftrightarrow\left[{}\begin{matrix}m=2\left(tm\right)\\m=-1\left(tm\right)\end{matrix}\right.\)

 Vì \(x_2\)là nghiệm của phương trình

=> \(x_2^2-5x_2+3=0\)

=> \(x_2+1=x^2_2-4x_2+4=\left(x_2-2\right)^2\)

Theo viet ta có

\(\hept{\begin{cases}x_1+x_2=5\\x_1x_2_{ }=3\end{cases}}\)=> \(x_1^2+x_2^2=19\)

Khi đó

\(A=||x_1-2|-|x_2-2||\)

=> \(A^2=\left(x^2_1+x_2^2\right)-4\left(x_1+x_2\right)+8-2|\left(x_1-2\right)\left(x_2-2\right)|\)

=> \(A^2=19-4.5+8-2|3-2.5+4|=1\)

Mà A>0(đề bài)

=> A=1

Vậy A=1

dcv_new 

dcv - new

Thay m = - 1 vào thì ta có: \(x^2-x-6=0\)

<=> x = 3 hoặc x = -2 

Vậy m = -1 và x2 = - 2

a, Thay \(x_1=3\)vào phương trình , khi đó :

\(pt< =>\)\(3^2+3m+2m-4=0\)

\(< =>5m+5=0\)

\(< =>m=-\frac{5}{5}=-1\)

Thay \(m=-1\)vào phương trình , khi đó :

\(pt< =>x^2-x+2=0\)

\(< =>x=\varnothing\left(vo-nghiem\right)\)(giải delta)

Vậy phương trình chỉ có nghiệm kép khi \(m=-1\)

b, Theo hệ thức vi ét ta có : \(\hept{\begin{cases}x_1+x_2=-\frac{b}{a}=-m\\x_1x_2=\frac{c}{a}=2m-4\end{cases}}\)

Khi đó \(A=\frac{2m-4+3}{-m}=\frac{2m-1}{-m}\)

Bạn thiếu đề rồi thì phải !

em bị nhầm ạ =(( 

1.

\(a+b+c=0\) nên pt luôn có 2 nghiệm

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

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

\(A=\dfrac{m^2+2-\left(m^2-2m+1\right)}{m^2+2}=1-\dfrac{\left(m-1\right)^2}{m^2+2}\le1\)

Dấu "=" xảy ra khi \(m=1\)

2.

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

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

\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

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

\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)

B=\(\frac{4m+1}{4m^2+2}\)

B=\(\frac{4m+1}{4m^2+2}\)

\(\Delta'=\left[-\left(m+4\right)\right]^2-1\left(m^2-8\right)=m^2+8m+16-m^2+8=8m+24\)

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow8m+24\ge0\Leftrightarrow m\ge-3\)

Áp dụng định lý Vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=2m+8\\x_1x_2=m^2-8\end{matrix}\right.\)

\(A=x^2_1+x^2_2-x_1-x_2\\ =\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)\\ =\left(2m+8\right)^2-2\left(m^2-8\right)-\left(2m+8\right)\\ =4m^2+32m+64-2m^2+16-2m-16\\ =2m^2+30m+64\)

A_min=\(-\dfrac{97}{2}\)\(\Leftrightarrow m=-\dfrac{15}{2}\)

\(B=x^2_1+x^2_2-x_1x_2\\ =\left(x_1+x_2\right)^2-3x_1x_2\\ =\left(2m+8\right)^2-3\left(m^2-8\right)\\ =4m^2+32m+64-3m^2+24\\ =m^2+32m+88\)

B_min=-168\(\Leftrightarrow\)m=-16

WTFast.

a: \(x_1+x_2=-\dfrac{b}{a}=2\sqrt{3};x_1\cdot x_2=\dfrac{c}{a}=1\)

Đặt \(A=x_1-x_2\)

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

\(=\left(2\sqrt{3}\right)^2-4\cdot1=12-4=8\)

=>\(\left[{}\begin{matrix}x_1-x_2=2\sqrt{2}\\x_1-x_2=-2\sqrt{2}\end{matrix}\right.\)

b: \(\dfrac{3x_1^2+5x_1x_2+3x_2^2}{4x_1^3\cdot x_2+4x_1\cdot x_2^3}\)

\(=\dfrac{3\left(x_1^2+x_2^2\right)+5x_1x_2}{4x_1x_2\left(x_1^2+x_2^2\right)}\)

\(=\dfrac{3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+5x_1x_2}{4x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)

\(=\dfrac{3\left(x_1+x_2\right)^2-x_1x_2}{4x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)

\(=\dfrac{3\cdot12-1}{4\cdot1\cdot\left[12-2\cdot1\right]}=\dfrac{35}{4\cdot10}=\dfrac{7}{8}\)

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