1, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=-6\end{matrix}\right.\)

\(A=\left(x_1-2x_2\right)\left(2x_1-x_2\right)\\ =2x_1^2-4x_1x_2-x_1x_2+2x_1^2\\ =2\left(x_1^2+x_2^2\right)-5x_1x_2\\ =2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2\\ =2\left(-5\right)^2-4.\left(-6\right)-5.\left(-6\right)\\ =104\)

2, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=-3\end{matrix}\right.\)

\(B=x_1^3x_2+x_1x_2^3\\ =x_1x_2\left(x_1^2+x_2^2\right)\\ =\left(-3\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\left(-3\right)\left[5^2-2\left(-3\right)\right]\\ =-93\)

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

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

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

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

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

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

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

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

a)

\(m=6\)

\(\Rightarrow x^2+5x+6=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

b)

\(\left|x_1-x_2\right|=3\)

\(\Leftrightarrow\left(x_1-x_2\right)^2=9\)

\(\Leftrightarrow x_1^2=2x_1x_2+x^2_2=9\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=9\)

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

\(\Rightarrow25-4m=9\)

\(\Leftrightarrow4m=16\)

\(\Leftrightarrow m=4\)

`1)`

$a\big)\Delta=7^2-5.4.1=29>0\to$ PT có 2 nghiệm pb

$b\big)$

Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{7}{5}\\x_1x_2=\dfrac{1}{5}\end{matrix}\right.\)

\(A=\left(x_1-\dfrac{7}{5}\right)x_1+\dfrac{1}{25x_2^2}+x_2^2\\ \Rightarrow A=\left(x_1-x_1-x_2\right)x_1+\left(\dfrac{1}{5}\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\\ \Rightarrow A=-x_1x_2+\left(x_1x_2\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\)

\(\Rightarrow A=-x_1x_2+x_1^2+x_2^2\\ \Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2\\ \Rightarrow A=\left(\dfrac{7}{5}\right)^2-3\cdot\dfrac{1}{5}=\dfrac{34}{25}\)

a, - Thay m = 6 vào phương trình ta được : \(x^2-5x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy ...

b, - Xét phương trình trên có : \(\Delta=b^2-4ac=25-4m\)

- Để phương trình có 2 nghiệm phân biệt <=> \(m< \dfrac{25}{4}\)

- Theo viet ta có : \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m\end{matrix}\right.\)

- Ta có : \(\left|x_1-x_2\right|=3\)

\(\Leftrightarrow x^2_1+x^2_2-2\left|x_1x_2\right|=\left(x_1+x_2\right)^2-2\left(x_1x_2+\left|x_1x_2\right|\right)=9\)

\(\Leftrightarrow m+\left|m\right|=8\)

\(\Leftrightarrow2m=8\)

\(\Leftrightarrow m=4\)

Vậy ...

a. thay m=-4 vào (1) ta có:

\(x^2-5x-6=0\)

Δ=b\(^2\)-4ac= (-5)\(^2\) - 4.1.(-6)= 25 + 24= 49 > 0

\(\sqrt{\Delta}=\sqrt{49}=7\)

x\(_1\)=\(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+7}{2}\)=6

x\(_2\)=\(\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-7}{2}\)=-1

vậy khi x=-4 thì pt đã cho có 2 nghiệm x\(_1\)=6; x\(_2\)=-1

Sửa đề: \(x_2^2-x_1^2=2\)

Ta có: \(\Delta=\left[-\left(m-3\right)\right]^2-4\cdot1\cdot\left(-2m+2\right)\)

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

\(=m^2-6m+9+8m-8\)

\(=m^2+2m+1\)

\(=\left(m+1\right)^2\ge0\forall m\)

Do đó: Phương trình luôn có hai nghiệm với mọi m

Áp dụng hệ thức Vi-et, ta được:

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

Ta có: \(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4\cdot x_1x_2\)

\(\Leftrightarrow\left(x_1-x_2\right)^2=\left(m-3\right)^2-4\left(-2m+2\right)\)

\(\Leftrightarrow\left(x_1-x_2\right)^2=m^2-6m+9+8m-8=m^2-2m+1\)

\(\Leftrightarrow x_1-x_2=m-1\)

Ta có: \(x_2^2-x_1^2=2\)

\(\Leftrightarrow\left(x_2-x_1\right)\left(x_2+x_1\right)=2\)

\(\Leftrightarrow\left(1-m\right)\left(m-3\right)=2\)

\(\Leftrightarrow m-3-m^2+3m-2=0\)

\(\Leftrightarrow-m^2+4m-5=0\)

\(\Leftrightarrow m^2-4m+5=0\)(Vô lý)

Vậy: Không có giá trị nào của m để phương trình có hai nghiệm thỏa mãn \(x_2^2-x_1^2=2\)

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

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

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

\(\Leftrightarrow\left(m-3\right)x_2+2m-2-x_1=2\)

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

\(\Leftrightarrow\left(m-2\right)x_2-m+3+2m-4=0\)

\(\Leftrightarrow\left(m-2\right)x_2=-m+1\Rightarrow x_2=\dfrac{-m+1}{m-2}\)

\(\Rightarrow x_1=m-3-x_2=\dfrac{m^2-4m+5}{m-2}\)

Thế vào \(x_1x_2=-2m+2\)

\(\Rightarrow\left(\dfrac{-m+1}{m-2}\right)\left(\dfrac{m^2-4m+5}{m-2}\right)=-2m+2\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\\dfrac{m^2-4m+5}{\left(m-2\right)^2}=2\left(1\right)\end{matrix}\right.\)

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

\(\Leftrightarrow m^2-4m+3=0\Rightarrow\left[{}\begin{matrix}m=1\\m=3\end{matrix}\right.\)

Đây chắc chắn là 1 đề bài sai (pt giải ra m là 1 pt bậc 3 hệ số xấu)

Bạn kiểm tra kĩ lại đề bài, phần hệ số các ẩn của pt bậc 2 ấy