\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)

\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)

\(=cos3a-cos3a\left(1-sin3a\right)\)

\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)

\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)

Vì a=\(\dfrac{5\pi}{6}\) nên: \(3a=\dfrac{5\pi}{2}\) => \(\cos3a=0\)

\(\pi-3a=\pi-\dfrac{5\pi}{2}=\dfrac{-3\pi}{2}\)

=> \(\cos\left(\pi-3a\right)=0\)

câu 1) ta có : \(M=\left(x^2-x\right)^2+\left(2x-1\right)^2=x^4-2x^3+x^2+4x^2-4x+1\)

\(=\left(x^2-x+2\right)^2-3=\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right)^2-3\)

\(\Rightarrow\dfrac{1}{16}\le M\le61\)

\(\Rightarrow M_{min}=\dfrac{1}{16}\)khi \(x=\dfrac{1}{2}\) ; \(M_{max}=61\) khi \(x=3\)

câu 2) điều kiện xác định : \(0\le x\le2\)
đặt \(\sqrt{2x-x^2}=t\left(t\ge0\right)\)

\(\Rightarrow M=-t^2+4t+3=-\left(t-2\right)^2+7\)

\(\Rightarrow3\le M\le7\)

câu 3) ta có : \(M=\left(x-2\right)^2+6\left|x-2\right|-6\ge-6\)

\(\Rightarrow M_{min}=-6\) khi \(x=2\)

4) điều kiện xác định \(-6\le x\le10\)

ta có : \(M=5\sqrt{x+6}+2\sqrt{10-x}-2\)

áp dụng bunhiacopxki dạng căn ta có :

\(-\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\le5\sqrt{x+6}+2\sqrt{10-x}\le\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\)

\(\Leftrightarrow-4\sqrt{29}\le5\sqrt{x+6}+2\sqrt{10-x}\le4\sqrt{29}\)

\(\Rightarrow-2-4\sqrt{29}\le B\le-2+4\sqrt{29}\)

\(\Rightarrow M_{max}=-2+4\sqrt{29}\) khi \(\dfrac{\sqrt{x+6}}{5}=\dfrac{\sqrt{10-x}}{2}\Leftrightarrow x=\dfrac{226}{29}\)

Ta có:

\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}=\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}=\frac{1}{5}\)

Chia tử và mẫu cho cosa ta có:

B=\(\dfrac{4\tan a+5}{2\tan a-3}\). Vì \(\cot a=\dfrac{1}{2}\) nên \(\tan a=2\)

=> B=13

chia tử và mẫu của B cho sina khác 0\(B=\dfrac{4\dfrac{sina}{sina}+5\dfrac{cosa}{sina}}{2\dfrac{sina}{sina}-3\dfrac{cosa}{sina}}=\dfrac{4+5cota}{2-3cota}=\dfrac{4+5\dfrac{1}{2}}{2-3\dfrac{1}{2}}=13\)

vay B = 13

a)
\(2sin30+3sin45^o-sin60^o=2.\dfrac{1}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}\)\(=\dfrac{2+3\sqrt{2}-\sqrt{3}}{2}\).
b)\(2cos30^o+3sin45^o-cos60^o=2.\dfrac{\sqrt{3}}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)\(=\dfrac{2\sqrt{3}+3\sqrt{2}-1}{2}\).

Ta có: \(\sqrt{x^2+y^2+4x-2y+5}+\sqrt{x^2+y^2-8x-14y+65}=6\sqrt{2}\)

\(\Leftrightarrow\sqrt{\left(x+2\right)^2+\left(y-1\right)^2}+\sqrt{\left(4-x\right)^2+\left(7-y\right)^2}=6\sqrt{2}\left(^∗\right)\)

Xét hai vectơ \(\overrightarrow{u}=\left(x+2;y-1\right)\)và \(\overrightarrow{v}=\left(4-x;7-y\right)\)

Ta có: \(\overrightarrow{u}+\overrightarrow{v}=\left(6;6\right)\Rightarrow\left|\overrightarrow{u}+\overrightarrow{v}\right|=\sqrt{6^2+6^2}=6\sqrt{2}\)

Do vậy \(\left(^∗\right)\)trở thành\(\overrightarrow{u}+\overrightarrow{v}=\left|\overrightarrow{u}+\overrightarrow{v}\right|\)

Điều này xảy ra khi và chỉ khi \(\overrightarrow{u}\)và \(\overrightarrow{v}\)cùng hướng

\(\Leftrightarrow\hept{\begin{cases}\left(x+2\right)\left(7-y\right)=\left(y-1\right)\left(4-x\right)\\\left(x+2\right)\left(4-x\right)\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=x+3\\-2\le x\le4\end{cases}}\)

Khi y = x + 3 thì \(x^2+y^2-2x+2y+2=2x^2+6x+17\)

Xét hàm số \(f\left(x\right)=2x^2+6x+17\)trên đoạn \(\left[-2;4\right]\)

Ta có: \(-\frac{6}{2.2}=\frac{-3}{2}\in\left[-2;4\right]\)và \(f\left(-2\right)=13;f\left(-\frac{3}{2}\right)=\frac{25}{2};f\left(4\right)=73\)

Suy ra \(|^{min}_{\left[-2;4\right]}f\left(x\right)=\frac{25}{2}\);\(|^{max}_{\left[-2;4\right]}f\left(x\right)=73\)

Do đó \(m=\frac{25}{2};M=73\)và \(n+M=\frac{171}{2}\)

Vậy \(n+M=\frac{171}{2}\)