d: Ta có: \(\text{Δ}=\left(m+1\right)^2-4\cdot2\cdot\left(m+3\right)\)

\(=m^2+2m+1-8m-24\)

\(=m^2-6m-23\)

\(=m^2-6m+9-32\)

\(=\left(m-3\right)^2-32\)

Để phương trình có hai nghiệm phân biệt thì \(\left(m-3\right)^2>32\)

\(\Leftrightarrow\left[{}\begin{matrix}m-3>4\sqrt{2}\\m-3< -4\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m>4\sqrt{2}+3\\m< -4\sqrt{2}+3\end{matrix}\right.\)

Áp dụng hệ thức Vi-et, ta được:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{m+1}{2}\\x_1x_2=\dfrac{m+3}{2}\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{m+1}{2}\\x_1-x_2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x_1=\dfrac{m+3}{2}\\x_2=x_1-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{m+3}{4}\\x_2=\dfrac{m+3}{4}-\dfrac{4}{4}=\dfrac{m-1}{4}\end{matrix}\right.\)

Ta có: \(x_1x_2=\dfrac{m+3}{2}\)

\(\Leftrightarrow\dfrac{\left(m+3\right)\left(m-1\right)}{16}=\dfrac{m+3}{2}\)

\(\Leftrightarrow\left(m+3\right)\left(m-1\right)=8\left(m+3\right)\)

\(\Leftrightarrow\left(m+3\right)\left(m-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-3\\m=9\end{matrix}\right.\)

b) phương trình có 2 nghiệm  \(\Leftrightarrow\Delta'\ge0\)

\(\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m+3\right)\ge0\)

\(\Leftrightarrow m^2-2m+1-m^2-3m+m+3\ge0\)

\(\Leftrightarrow-4m+4\ge0\)

\(\Leftrightarrow m\le1\)

Ta có: \(x_1^2+x_1x_2+x_2^2=1\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=1\)

Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)

\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)

\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)

\(\Leftrightarrow4m^2-10m-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)

Câu c:

(a) Khi \(m=2,\left(1\right)\Leftrightarrow x^2-4x-5=0\left(2\right)\).

Phương trình (2) có \(a-b+c=1-\left(-4\right)+\left(-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{c}{a}=5\end{matrix}\right.\).

Vậy: Khi \(m=2,S=\left\{-1;5\right\}\).

(b) Điều kiện: \(x_1,x_2\ne0\Rightarrow m\in R\)

Phương trình có nghiệm khi:

\(\Delta'=\left(-m\right)^2-1\cdot\left(-m^2-1\right)\ge0\)

\(\Leftrightarrow2m^2+1\ge0\left(LĐ\right)\)

Suy ra, phương trình (1) có nghiệm với mọi \(m\).

Theo định lí Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=-m^2-1\end{matrix}\right.\)

Theo đề: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-\dfrac{5}{2}\)

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

\(\Leftrightarrow2\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow2\left(2m\right)^2+\left(-m^2-1\right)=0\)

\(\Leftrightarrow7m^2=1\Leftrightarrow m=\pm\dfrac{\sqrt{7}}{7}\) (thỏa mãn).

Vậy: \(m=\pm\dfrac{\sqrt{7}}{7}.\)

a, bạn tự làm 

b, Thay x = 3 vào pt trên ta được 

\(9-3m-3=0\Leftrightarrow6-3m=0\Leftrightarrow m=2\)

Thay m = 2 vào ta được \(x^2-2x-3=0\)

Ta có a - b + c = 1 + 2 - 3 = 0 

vậy pt có 2 nghiệm x = -1 ; x = 3 

c, \(\Delta=m^2-4\left(-3\right)=m^2+12>0\)

vậy pt luôn có 2 nghiệm pb 

\(x_1x_2+5\left(x_1+x_2\right)-1997=0\)

\(\Rightarrow-3+5m-1997=0\Leftrightarrow5m-2000=0\Leftrightarrow m=400\)

Đề bài sai bạn

Biểu thức \(\left|\dfrac{x_1+x_2+4}{x_1+x_2}\right|=\left|1+\dfrac{1}{m}\right|\)  này ko tồn tại max, chỉ tồn tại min

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

\(\Rightarrow25-4m=9\)

\(\Leftrightarrow4m=16\)

\(\Leftrightarrow m=4\)

Đăng lại lớp đi chụy :)

\(\Delta'=\left(-1\right)^2-\left(m-1\right)=1-m+1=2-m\)

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)

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

\(x^4_1-x^3_1=x^4_2-x^3_2\\ \Leftrightarrow\left(x^4_1-x_2^4\right)-\left(x^3_1+x^3_2\right)=0\\ \Leftrightarrow\left(x^2_1-x^2_2\right)\left(x^2_1+x^2_2\right)-\left(x_1+x_2\right)\left(x^2_1+x^2_2-x_1x_2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=0\\ \Leftrightarrow\left(m-1\right).2\left[2^2-2\left(m-1\right)\right]-2\left[2^2-3\left(m-1\right)\right]=0\)

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

\(\Leftrightarrow\left(2m-2\right)\left(6-2m\right)-2\left(7-3m\right)=0\)

\(\Leftrightarrow...\)

\(\Delta'=\left(-1\right)^2-\left(m-1\right)=1-m+1=2-m\)

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)

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

\(x^4_1-x^3_1=x^4_2-x^3_2\\ \Leftrightarrow\left(x^4_1-x_2^4\right)-\left(x^3_1-x^3_2\right)=0\\ \Leftrightarrow\left(x^2_1-x^2_2\right)\left(x^2_1+x^2_2\right)-\left(x_1-x_2\right)\left(x^2_1+x^2_2+x_1x_2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]=0\)

\(\Leftrightarrow\left(x_1-x_2\right).2\left(4-2m+2\right)-\left(x_1-x_2\right)\left(4-m+1\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right).2\left(6-2m\right)-\left(x_1-x_2\right)\left(5-m\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(12-4m-5+m\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(7-3m\right)=0\)

\(\Leftrightarrow...\)

\(\Delta'=9-6m+m^2=\left(m-3\right)^2\ge0;\forall m\)

\(\Rightarrow\) Pt luôn có nghiệm với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-6\\x_1x_2=6m-m^2\end{matrix}\right.\)

Do \(x_1\) là nghiệm nên: \(x_1^2+6x_1+6m-m^2=0\Leftrightarrow2x_1^2+12x_1=2m^2-12\)

\(x_1^3-x_2^3+2x_1^2+12x_1+72=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]+2m^2-12m+72=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(m^2-6m+36\right)+2m^2-12m+72=0\)

\(\Leftrightarrow\left(x_1-x_2+2\right)\left(m^2-6m+36\right)=0\)

\(\Leftrightarrow x_1-x_2+2=0\) (do \(m^2-6m+36=\left(m-3\right)^2+27>0;\forall m\))

Kết hợp với \(x_1+x_2=-6\) ta được: 

\(\left\{{}\begin{matrix}x_1-x_2=-2\\x_1+x_2=-6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-4\\x_2=-2\end{matrix}\right.\)

Thế vào \(x_1x_2=6m-m^2\)

\(\Rightarrow8=6m-m^2\Rightarrow m^2-6m+8=0\Rightarrow\left[{}\begin{matrix}m=2\\m=4\end{matrix}\right.\)