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

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

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

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

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

\(x^2-2\left(m+4\right)x+m^2+8m-9=0\left(1\right)\)

Ta giải \(\Delta=[-2\left(m+4\right)]^2-4\left(m^2+8m-9\right)=100>0\forall m\)

suy ra pt có 2 nghiệm phân biệt \(x_1,x_2\forall m\).

Ta có: \(x_1=m-1\), \(x_2=m+1\) (thay \(\Delta\) vào công thức tìm nghiệm phân biệt).

Gọi \(A=\dfrac{x_1^2+x_2^2-48}{x_1^2+x_2^2}\).

\(\Rightarrow A=1-\dfrac{48}{x_1^2+x_2^2}=1-\dfrac{48}{\left(m-1\right)^2+\left(m+1\right)^2}=1-\dfrac{24}{m^2+1}\).

Để biểu thức A nguyên thì \(\dfrac{24}{m^2+1}\) nguyên, suy ra \(m^2+1\inƯ\left(24\right)\).

\(\Rightarrow m^2+1\in\left\{1;2;4;6;8;12;24\right\}\)

\(\Rightarrow m\in\left\{0;\pm1\right\}\) (vì m nhận giá trị nguyên)

Vậy \(m\in\left\{0;\pm1\right\}\) là giá trị cần tìm.

Mình chỉnh sửa lại một chút nhé.

\(A=1-\dfrac{24}{m^2+2}\)

\(\Rightarrow...\)\(\Rightarrow\)\(m^2+2\in\left\{1;2;3;4;6;8;12;24\right\}\)

\(\Rightarrow m\in\left\{0;\pm1;\pm2\right\}\)

Vậy...

\(x^{2_2}-x^{2_1}=?\)

Bổ sung thêm vào bạn nhé

Lời giải:
1. 

Khi $m=-1$ thì pt trở thành: $x^2+4x+2=0$

$\Leftrightarrow (x+2)^2=2$

$\Leftrightarrow x+2=\pm \sqrt{2}$

$\Leftrightarrow x=-2\pm \sqrt{2}$

2.

Ta thấy: $\Delta'=(m-1)^2+2m=m^2+1>0$ với mọi $m\in\mathbb{R}$

Do đó pt luôn có 2 nghiệm pb với mọi $m$

Áp dụng định lý Viet:

$x_1+x_2=2(m-1)$

$x_1x_2=-2m$

Khi đó:

$x_1^2+x_1-x_2=5-2m=3-2(m-1)=3-x_1-x_2$

$\Leftrightarrow x_1^2+2x_1-3=0$

$\Leftrightarrow (x_1-1)(x_1+3)=0$

$\Leftrightarrow x_1=1$ hoặc $x_1=-3$

Nếu $x_1=1$

$\Leftrightarrow x_2+1=2m-2$ và $x_2=-2m$

$\Rightarrow 2x_2+1=-2$

$\Leftrightarrow x_2=\frac{-3}{2}$

$-2m=x_1x_2=\frac{-3}{2}$

$m=\frac{3}{4}$

-------------

Nếu $x_1=-3$

$\Leftrightarrow x_2-3=2m-2$ và $-3x_2=-2m$

$\Leftrightarrow m=\frac{-3}{4}$

1, ĐKXĐ:\(x\ne2,y\ne1\)

Đặt `1/(x-2)` = a, `1/(y-1)` = b

\(Hệ.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\\b=\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{y-1}=\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\3y-3=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\\y=\dfrac{8}{3}\end{matrix}\right.\)\(2,\Delta'=\left[-\left(m+1\right)\right]^2-4m=m^2+2m+1-4m=m^2-2m+1=\left(m-1\right)^2\ge0\)

Để pt có 2 nghiệm phân biệt thì \(\Delta'>0\Leftrightarrow\left(m-1\right)^2>0\Leftrightarrow m-1\ne0\Leftrightarrow m\ne1\)

b, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=4m\end{matrix}\right.\)

\(\left(x_1-x_2\right)^2-x_1x_2=3\\ \Leftrightarrow\left(x_1+x_2\right)^2-5x_1x_2=3\\ \Leftrightarrow\left(2m+2\right)^2-5.4m-3=0\\ \Leftrightarrow4m^2+8m+4-20m-3=0\\ \Leftrightarrow4m^2-12m+1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+2\sqrt{2}}{2}\\x=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\)

`a)` Ptr có:`\Delta' =[-(m-1)]^2-(-3-m)`

                            `=m^2-2m+1+3+2m=m^2+4 > 0 AA m`

  `=>` Ptr có `2` nghiệm `AA m`

`b) AA m`, áp dụng Vi-ét có:`{(x_1+x_2=[-b]/a=2m-2),(x_1.x_2=c/a=-3-m):}`

Ta có:`x_1 ^2+x_2 ^2 >= 10`

`<=>(x_1+x_2)^2-2x_1.x_2 >= 10`

`<=>(2m-2)^2-2(-3-m) >= 10`

`<=>4m^2-8m+4+6+2m >= 10`

`<=>4m^2-6m+10 >= 10`

`<=>4m^2-6m >= 0`

`<=>2m(2m-3) >= 0`

`<=>` $\left[\begin{matrix} m \ge \dfrac{3}{2}\\ m \le 0\end{matrix}\right.$

Vậy `m >= 3/2` hoặc `m <= 0` thì t/m yêu cầu đề bài

a: \(\text{Δ}=\left(2m-2\right)^2-4\left(-m-3\right)\)

\(=4m^2-8m+4+4m+12=4m^2-4m+16\)

\(=\left(2m-1\right)^2+15>0\)

Do đó: Phương trình luôn có hai nghiệm

b: Theo Vi-et, ta được:

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

Ta có: \(x_1^2+x_2^2>=10\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2>=10\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(-m-3\right)>=10\)

\(\Leftrightarrow4m^2-8m+4+2m+6-10>=0\)

\(\Leftrightarrow4m^2-6m>=0\)

=>2m(2m-3)>=0

=>m>=3/2 hoặc m<=0

a, Ta có:

\(\Delta'=\left[-\left(m-1\right)\right]^2-1\left(-3-m\right)\\ =\left(m-1\right)^2-\left(-3-m\right)\\ =m^2-2m+1+3+m\\ =m^2-m+4\\ =\left(m^2-m+\dfrac{1}{4}\right)+\dfrac{15}{4}\\ =\left(m-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)

Suy ra pt luôn có 2 nghiệm phân biệt

b, Theo Vi-ét:\(x_1+x_2=2m-2;x_1x_2=-m-3\)

\(x_1^2+x_2^2\ge10\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2\ge10\\ \Leftrightarrow\left(2m-2\right)^2-2\left(-m-3\right)-10\ge0\\ \Leftrightarrow4m^2-8m+4+2m+6-10\ge0\\ \Leftrightarrow4m^2-6m\ge0\\ \Leftrightarrow2m\left(2m-3\right)\ge0\\ \Leftrightarrow m\left(2m-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m\ge0\\2m-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}m\le0\\2m-3\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m\ge0\\m\ge\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}m\le0\\m\le\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m\ge\dfrac{3}{2}\\m\le0\end{matrix}\right.\)

1: \(\text{Δ}=\left(2m-2\right)^2-4\left(m-3\right)\)

\(=4m^2-8m+4-4m+12\)

\(=4m^2-12m+16\)

\(=\left(2m-3\right)^2+7>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

2: Theo vi-et, ta được:

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

Ta có: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=x_1x_2\)

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

\(\Leftrightarrow\left(2m-2\right)^2-2\left(m-3\right)-\left(m-3\right)^2=0\)

\(\Leftrightarrow4m^2-16m+4-2m+6-m^2+6m-9=0\)

\(\Leftrightarrow3m^2-12m+1=0\)

\(\text{Δ}=\left(-12\right)^2-4\cdot3\cdot1=144-12=132>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{12-2\sqrt{33}}{6}=\dfrac{6-\sqrt{33}}{3}\\x_2=\dfrac{6+\sqrt{33}}{3}\end{matrix}\right.\)

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

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

\(\Delta=\left(m-2\right)^2+8>0\) với mọi m . Vậy pt có 2 nghiệm phân biệt với mọi m 

Do : \(x_1x_2=-8\) nên \(x_2=\dfrac{-8}{x1}\)

\(Q=\left(x_1^2-1\right)\left(x_2^2-4\right)=\left(x_1^2-1\right)\left(\dfrac{64}{x_1^2}-4\right)=68-4\left(x_1^2+\dfrac{16}{x_1^2}\right)\le68-4.8=36\)

\(\left(x_1^2+\dfrac{16}{x_1^2}\ge8\right)\)\(;Q=36\) khi và chỉ khi x₁ = ( 2 ; -2 )