\(x^2=mx-m+5\Leftrightarrow x^2-mx+m-5=0\left(1\right)\)

\(\Delta>0\Leftrightarrow m^2-4\left(m-5\right)>0\Leftrightarrow m^2-4m+20>0\left(đúng\forall m\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x1+x2=m\\x1x2=m-5\end{matrix}\right.\)

\(\dfrac{1}{x1}+\dfrac{1}{x2}=-\dfrac{3}{2}\left(x1;x2\ne0\right)\Rightarrow\dfrac{x1+x2}{x1x2}=-\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{m}{m-5}=-\dfrac{3}{2}\Leftrightarrow m=3\) \(m=3\Rightarrow x_{12}=\dfrac{3\pm\sqrt{17}}{2}\left(tm\right)\Rightarrow m=3\left(tm\right)\)

Đề thiếu dữ kiện liên quan đến ABC. Bạn coi lại.

\(A=(-\infty;4]\)

\(B=\left[3m+2;+\infty\right]\)

Để A giao B bằng rỗng thì  3m+2>4

hay m>2/3

Ta có :

\(A=\left(-\infty;4\right)\)

\(B=\left[3m+2;+\infty\right]\)

\(\Rightarrow\) A giao với B = rỗng thì : \(3m+2>4\),hay :\(m>\dfrac{2}{3}\)

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

\(=-x^2+2x+3\)

\(\Delta'=1^2-1\cdot\left(-1\right)\cdot3=4>0,a=-1< 0\)

\(\Rightarrow f\left(x\right)\) có hai nghiệm phân biệt : \(x_1=-1,x_2=3\)

Khi đó \(f\left(x\right)>0\forall x\in\left(-\infty;-1\right)\cup\left(3;+\infty\right)\)

\(f\left(x\right)< 0\forall x\in\left(-1;3\right)\).

\(\cos^2a=1-\dfrac{25}{169}=\dfrac{144}{169}\)

mà \(\cos a< 0\)

nên \(\cos a=-\dfrac{12}{13}\)

\(\sin\left(\text{Δ}+\dfrac{\Pi}{3}\right)=\sin\text{Δ}\cdot\dfrac{1}{2}+\cos\text{Δ}\cdot\dfrac{\sqrt{3}}{2}\)

\(=\dfrac{-5}{13}\cdot\dfrac{1}{2}+\dfrac{-12}{13}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{-5-12\sqrt{3}}{26}\)

Có :

\(cos^2a=1-\dfrac{25}{169}=\dfrac{144}{169}\)

-> Mà cos a < 0 , nên cos a = \(-\dfrac{12}{13}\)

Ta có :

\(sin\left(\Delta+\dfrac{11}{3}\right)=sin\Delta\times\dfrac{1}{2}+cos\Delta\times\dfrac{\sqrt{3}}{2}\)

Vậy :

\(\dfrac{-5}{13}\times\dfrac{1}{2}+\dfrac{-12}{13}\times\dfrac{\sqrt{3}}{2}=\dfrac{-5-12\sqrt{3}}{26}\)