\(a,\)\(A=\left\{x\in R|x< 3\right\}\Rightarrow A=\left(\text{ -∞;3}\right)\)

\(B=\left\{-1;0;1;2;3;4;5\right\}\)

\(\Rightarrow A\cap B=\left\{-1;0;1;2\right\}\)

\(b,x=-1\Rightarrow y=1-2\left(-1\right)+m=m+3\) 

\(x=1\Rightarrow y=1-2+m=m-1\)

\(\Rightarrow C=(m-1;m+3]\subset A\)

\(\Rightarrow C\subset A\Leftrightarrow m+3< 3\Leftrightarrow m< 0\)

\(B\le\frac{x^2+25-x^2}{2}=\frac{25}{2}\)

\(\Rightarrow B_{max}=\frac{25}{2}\) khi \(\left|x\right|=\sqrt{25-x^2}\Leftrightarrow x=\pm\frac{5\sqrt{2}}{2}\)

Áp dụng BĐT Bunhiacopxki ta có :

\(\left(3\sqrt{x-1}+4\sqrt{5-x}\right)^2\le\left(3^2+4^2\right)\left(x-1+5-x\right)\)

\(\Leftrightarrow\left(3\sqrt{x-1}+4\sqrt{5-x}\right)^2\le100\)

\(\Leftrightarrow f\left(x\right)\le10\)

Dấu "=" xảy ra :

\(\Leftrightarrow\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}\)

Vậy...

\(-2x^2-4x+3-m=0\)

\(\Delta'=\left(-2\right)^2-\left(3-m\right).\left(-2\right)=10-2m\)

Để phương trình có nghiệm

\(\Leftrightarrow\Delta'\ge0\Leftrightarrow10-2m\ge0\Leftrightarrow m\le5\)

\(\Rightarrow A\)

\(sinx+cosx=m\Leftrightarrow\left(sinx+cosx\right)^2=m^2\)

\(\Leftrightarrow1+2sinx.cosx=m^2\Rightarrow sinx.cosx=\dfrac{m^2-1}{2}\)

\(A=sin^2x+cos^2x=1\)

\(B=sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)

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

\(C=\left(sin^2+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1-2\left(\dfrac{m^2-1}{2}\right)^2\)

\(D=\left(sin^2x\right)^3+\left(cos^2x\right)^3=\left(sin^2x+cos^2x\right)^3-3\left(sin^2x+cos^2x\right)\left(sinx.cosx\right)^2\)

\(=1-3\left(\dfrac{m^2-1}{2}\right)^2\)

A. \(\sin A = \sin \,(B + C)\)

Ta có: \((\widehat A  + \widehat C) + \widehat B= {180^o}\)

\(\Rightarrow \sin \,(B + C) = \sin A\)

=> A đúng.

B. \(\cos A = \cos \,(B + C)\)

Sai vì \(\cos \,(B + C) =  - \cos A\)

C. \(\;\cos A > 0\) Không đủ dữ kiện để kết luận.

Nếu \({0^o} < \widehat A < {90^o}\) thì \(\cos A > 0\)

Nếu \({90^o} < \widehat A < {180^o}\) thì \(\cos A < 0\)

D. \(\sin A\,\, \le 0\)

Ta có \(S = \frac{1}{2}bc.\sin A > 0\). Mà \(b,c > 0\)

\( \Rightarrow \sin A > 0\)

=> D sai.

Chọn A

ta có \(3\le a\le5\)

=> (a-3)(a-5) ≤ 0

<=> a²-5a-3a+15 ≤ 0

<=> a²-8a+15 ≤ 0 (1)

\(3\le b\le5\)

=> (b-3)(b-5) ≤ 0

<=> b² -8b +15 ≤ 0 (2)

\(3\le c\le5\)

=> (c-3)(c-5) ≤ 0

<=> c² -8c +15 ≤ 0 (3)

(1)+(2)+ (3)

=> a²+b²+c² -8a-8b-8c +45 ≤ 0

<=> 50-8(a+b+c)+45 ≤ 0

<=> -8(a+b+c) ≤ -95

<=> a+b+c ≥ \(\dfrac{95}{8}\)

=> Min A= 95/8