Chọn A

+) Xét \(\beta  =  - \alpha \), khi đó:

\(\begin{array}{l}cos\beta  = cos\left( {-{\rm{ }}\alpha } \right) = cos\alpha ;\\sin\beta  = sin\left( {-{\rm{ }}\alpha } \right) = -sin\alpha  \Leftrightarrow sin\alpha  = -sin\beta .\end{array}\)

Do đó A thỏa mãn.

Đáp án: A

2.

ĐK: \(2x-y\ge0;y\ge0;y-x-1\ge0;y-3x+5\ge0\)

\(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\left(1\right)\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(1-y\right)\sqrt{2x-y}+y-1+2x-y-1-\left(2x-y-1\right)\sqrt{y}=0\)

\(\Leftrightarrow\left(1-y\right)\left(\sqrt{2x-y}-1\right)+\left(2x-y-1\right)\left(1-\sqrt{y}\right)=0\)

\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(1+\sqrt{y}\right)+\left(\sqrt{2x-y}-1\right)\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}+1\right)=0\)

\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(\sqrt{y}+\sqrt{2x-y}+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2x-1\end{matrix}\right.\) (Vì \(\sqrt{y}+\sqrt{2x-y}+2>0\))

Nếu \(y=1\), khi đó:

\(\left(1\right)\Leftrightarrow x-5=\sqrt{-x}+\sqrt{-3x+6}\)

Phương trình này vô nghiệm

Nếu \(y=2x-1\), khi đó:

\(\left(1\right)\Leftrightarrow2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\) (Điều kiện: \(2\le x\le4\))

\(\Leftrightarrow2x\left(x-3\right)+x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1\right)=0\)

Ta thấy: \(1+\sqrt{x-2}\ge1\Rightarrow-\dfrac{1}{1+\sqrt{x-2}}\ge-1\Rightarrow1-\dfrac{1}{1+\sqrt{x-2}}\ge0\)

Lại có: \(\dfrac{1}{1+\sqrt{4-x}}>0\); \(2x>0\)

\(\Rightarrow\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1>0\)

Nên phương trình \(\left(1\right)\) tương đương \(x-3=0\Leftrightarrow x=3\Rightarrow y=5\)

Ta thấy \(\left(x;y\right)=\left(3;5\right)\) thỏa mãn điều kiện ban đầu.

Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(3;5\right)\)

\(a)\;sin(\alpha  + \beta ).sin(\alpha  - \beta ) = \;\frac{1}{2}.\left[ {cos\left( {\alpha  + \beta  - \alpha  + \beta } \right) - cos\left( {\alpha  + \beta  + \alpha  - \beta } \right)} \right]\)

\(\begin{array}{l} = \;\frac{1}{2}.(cos2\beta  - cos2\alpha ) = \;\frac{1}{2}.(1 - 2si{n^2}\beta  - 1 + 2si{n^2}\alpha )\\ = si{n^2}\alpha  - si{n^2}\beta \end{array}\)

\(\begin{array}{l}b)\;co{s^4}\alpha  - co{s^4}\left( {\alpha  - \frac{\pi }{2}} \right) = \;co{s^4}\alpha  - si{n^4}\alpha \\ = \;(co{s^2}\alpha  + si{n^2}\alpha )(co{s^2}\alpha  - si{n^2}\alpha )\\ = \;co{s^2}\alpha -si{n^2}\alpha  = cos2\alpha .\end{array}\)

Theo Viet: \(\left\{{}\begin{matrix}tana+tanb=p\\tana.tanb=q\end{matrix}\right.\)

\(\Rightarrow tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}=\frac{p}{1-q}\)

\(\Rightarrow A=cos^2\left(a+b\right)+psin\left(a+b\right)+q.sin^2\left(a+b\right)\)

\(=\frac{1}{cos^2\left(a+b\right)}\left(1+p.\frac{sin\left(a+b\right)}{cos\left(a+b\right)}+q.\frac{sin^2\left(a+b\right)}{cos^2\left(a+b\right)}\right)\)

\(=\left[1+tan^2\left(a+b\right)\right]\left[1+p.tan\left(a+b\right)+q.tan^2\left(a+b\right)\right]\)

\(=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]\left[1+\frac{p^2}{1-q}+\frac{p^2q}{\left(1-q\right)^2}\right]\)

\(=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]\left[1+\frac{p^2}{\left(1-q\right)^2}\right]=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]^2\)

Ta có:

\(\begin{array}{l}\cos \alpha \cos \beta  = \cos \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2}\\ = \frac{1}{2}\left[ {\cos \left( {\frac{{\alpha  + \beta }}{2} + \frac{{\alpha  - \beta }}{2}} \right) + \cos \left( {\frac{{\alpha  + \beta }}{2} - \frac{{\alpha  - \beta }}{2}} \right)} \right]\\ = \frac{1}{2}\left( {\cos \alpha  + \cos \beta } \right)\end{array}\)

\(\begin{array}{l}\sin \alpha \sin \beta  = \sin \frac{{\alpha  + \beta }}{2}\sin \frac{{\alpha  - \beta }}{2}\\ = \frac{1}{2}\left[ {\cos \left( {\frac{{\alpha  + \beta }}{2} - \frac{{\alpha  - \beta }}{2}} \right) - \cos \left( {\frac{{\alpha  + \beta }}{2} + \frac{{\alpha  - \beta }}{2}} \right)} \right]\\ = \frac{1}{2}\left( {\cos \beta  - \cos \alpha } \right)\end{array}\)

\(\begin{array}{l}\sin \alpha \cos \beta  = \sin \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2}\\ = \frac{1}{2}\left[ {\sin \left( {\frac{{\alpha  + \beta }}{2} + \frac{{\alpha  - \beta }}{2}} \right) + \sin \left( {\frac{{\alpha  + \beta }}{2} - \frac{{\alpha  - \beta }}{2}} \right)} \right]\\ = \frac{1}{2}\left( {\sin \alpha  + \sin \beta } \right)\end{array}\)

Ta có: 

A. \(\alpha< \beta\)

\(\Rightarrow\left(0,3\right)^{\alpha}>\left(0,3\right)^{\beta}\)

Sai 

B. \(\alpha< \beta\)

\(\Rightarrow\pi^{\alpha}< \pi^{\beta}\)

Sai

C. \(\alpha< \beta\)

\(\Rightarrow\left(\sqrt{2}\right)^{\alpha}< \left(\sqrt{2}\right)^{\beta}\)

Đúng

D. \(\alpha< \beta\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{\alpha}>\left(\dfrac{1}{2}\right)^{\beta}\)

Sai 

⇒ Chọn C

a,

\(\begin{array}{l}\cos \left( {\alpha  - b} \right) + \cos \left( {\alpha  + \beta } \right)\\ = \cos \alpha \cos \beta  + \sin \alpha sin\beta  + \cos \alpha \cos \beta  - \sin \alpha sin\beta \\ = 2\cos \alpha \cos \beta \end{array}\)

\(\begin{array}{l}\cos \left( {\alpha  - b} \right) - \cos \left( {\alpha  + \beta } \right)\\ = \cos \alpha \cos \beta  + \sin \alpha sin\beta  - \cos \alpha \cos \beta  + \sin \alpha sin\beta \\ = 2\sin \alpha sin\beta \end{array}\)

b,

\(\begin{array}{l}\sin \left( {\alpha  - \beta } \right) - \sin \left( {\alpha  + \beta } \right)\\ = \sin \alpha \cos \beta  - \cos \alpha sin\beta  - \sin \alpha \cos \beta  - \cos \alpha sin\beta \\ =  - 2\cos \alpha sin\beta \end{array}\)

\(\begin{array}{l}\sin \left( {\alpha  - \beta } \right) + \sin \left( {\alpha  + \beta } \right)\\ = \sin \alpha \cos \beta  - \cos \alpha sin\beta  + \sin \alpha \cos \beta  + \cos \alpha sin\beta \\ = 2\sin \alpha \cos \beta \end{array}\)