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a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
\(cos\varphi=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{-1.2+3.1}{\sqrt{\left(-1\right)^2+3^2}.\sqrt{2^2+1^2}}=\frac{1}{5\sqrt{2}}\)
Lời giải:Áp dụng định lý cos ta có:
\(\cos A=\frac{AB^2+AC^2-BC^2}{2AB.AC}=\frac{-1}{2}\Rightarrow \widehat{A}=120^0\)
\(\cos B=\frac{BC^2+BA^2-AC^2}{2BC.BA}=\frac{-\sqrt{2}}{2}\Rightarrow \widehat{B}=45^0\)
\(\widehat{C}=180^0-(\widehat{A}+\widehat{B})=180^0-(120^0+45^0)=15^0\)
\(\widehat{ADB}=180^0-(\frac{\widehat{A}}{2}+\widehat{B})=180^0-(\frac{120^0}{2}+45^0)=75^0\)
Lời giải:
Theo BĐT Bunhiacopxky ta có:
$M^2=(\sin A+\sqrt{3}\cos A)^2\leq (\sin ^2A+\cos ^2A)(1+3)=1.4=4$
$\Rightarrow -2\leq M\leq 2$
Do đó $M$ không thể nhận giá trị $2\sqrt{3}$ vì $2\sqrt{3}>2$
Đáp án C.
\(A=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}-\dfrac{1}{\sqrt{a}-2}\)
=\(\dfrac{\left(\sqrt{a}+2\right).\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\left(\sqrt{a}-4\right).\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
Điều kiện bạn tự ghi nhé 
\(B=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}+\dfrac{\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\left(\sqrt{a}+3\right).\left(\sqrt{a}-3\right)-\left(\sqrt{a}-2\right).\left(\sqrt{a}+2\right)+\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{a-9-a+4+\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{1}{\sqrt{a}-2}\)
\(=\dfrac{1}{\sqrt{a}+1}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\)
a) \(2x-\dfrac{x-3}{5}-4x+1\le0\)
\(\Leftrightarrow10x-x+3-20x+5\le0\)
\(\Leftrightarrow-11x+8\le0\)
\(\Leftrightarrow x\ge\dfrac{8}{11}\)
\(\Rightarrow x\in\left(\dfrac{8}{11};+\infty\right)\)
b) \(\sqrt{x^2+2}\le x-1\)
\(\Leftrightarrow x^2+2\le x^2-2x+1\) \(\left(x-1\ge\sqrt{x^2+2}\ge\sqrt{2}\Rightarrow x\ge1+\sqrt{2}\right)\)
\(\Leftrightarrow x\le-\dfrac{1}{2}\)
\(\Rightarrow x\in\varnothing\)
c) \(\sqrt{x-1}+\sqrt{5-x}+\dfrac{1}{x-3}>\dfrac{1}{x-3}\) (\(x\in\left[1;5\right]\backslash\left\{3\right\}\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{5-x}>0\)
\(\Leftrightarrow4+2\sqrt{\left(x-1\right)\left(5-x\right)}>0\) ( luôn đúng )
vậy \(x\in\left[1;5\right]\backslash\left\{3\right\}\)