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 \(\left(x;y\right)=\left(3;5\right)\)

1.

\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)

\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)

\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)

\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)

\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)

\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)

\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)

\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)

2.

\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)

\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x\)

\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)

\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)

\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)

\(=tan^2a.cot^2b-2\)

3.

\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b\right).cosa-cos\left(a+b\right)sina\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b-a\right)\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sinb\)

b.

\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(2a+b\right)+sin\left(-b\right)=\dfrac{1}{2}sin\left(2a+b\right)+\dfrac{1}{2}sinb\)

\(\Leftrightarrow\dfrac{1}{2}sin\left(2a+b\right)=\dfrac{3}{2}sinb\)

\(\Leftrightarrow sin\left(2a+b\right)=3sinb\)

\(\dfrac{sin\left(a-b\right)}{sina.sinb}+\dfrac{sin\left(b-c\right)}{sinb.sinc}+\dfrac{sin\left(c-a\right)}{sinc.sina}\)

\(=\dfrac{sina.cosb-cosa.sinb}{sina.sinb}+\dfrac{sinb.cosc-cosb.sinc}{sinb.sinc}+\dfrac{sinc.cosa-cosc.sina}{sina.sinc}\)

\(=\dfrac{cosb}{sinb}-\dfrac{cosa}{sina}+\dfrac{cosc}{sincc}-\dfrac{cosb}{sinb}+\dfrac{cosa}{sina}-\dfrac{cosc}{sincc}\)

\(=0\)

nhường cho I don't know :))

Kiểm tra lại đề bài, \(cosa=\dfrac{1}{\sqrt{10}}\) hay \(cos\beta=\dfrac{1}{\sqrt{10}}\)?

a, \(\dfrac{1-sin2a}{1+sin2a}\)

\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)

\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)

\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)

\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)

\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)

b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)

\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)

\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)

\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)

\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)

\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)

Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)

Theo giả thiết ta có 3 góc: \(\alpha;\beta=\alpha+\dfrac{\pi}{3};\gamma=\alpha+\dfrac{2\pi}{3}\).
Ta có:
\(tan\alpha.tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{\pi}{3}\right).tan\left(\alpha+\dfrac{2\pi}{3}\right)+\)\(tan\left(\alpha+\dfrac{2\pi}{3}\right).tan\alpha\)
\(=tan\alpha\left[tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{2\pi}{3}\right)\right]\)\(+tan\left(a+\dfrac{\pi}{3}\right)tan\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=tan\alpha\dfrac{sin\left(2\alpha+\pi\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha+\dfrac{2\pi}{3}\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=tan\alpha\dfrac{-sin2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{cos\dfrac{\pi}{3}-cos\left(2\alpha+\pi\right)}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{-2sin^2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{\dfrac{1}{2}+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4sin^2\alpha+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4\left(1-cos^2\alpha\right)+2cos^2\alpha-1}{cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)}\)
\(=\dfrac{6cos^2\alpha-\dfrac{9}{2}}{\dfrac{1}{2}-cos2\alpha}\)
\(=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{1}{2}-\left(2cos^2\alpha-1\right)}=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{3}{2}-2cos^2\alpha}=-3\).

\(4cos\alpha.cos\beta cos\gamma=4cos\alpha cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(\dfrac{1}{2}-cos2\alpha\right)\)
\(=cos\alpha-2cos\alpha.cos2\alpha\)
\(=cos\alpha-\left(cos\alpha+cos3\alpha\right)\)
\(=-cos3\alpha\)
\(=cos\left(\pi+3\alpha\right)\)
\(=cos3\left(\dfrac{\pi}{3}+\alpha\right)\)
\(=cos3\beta\) (đpcm).

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

\(=2+2\left(cosa.cosb+sina.sinb\right)\)

\(=2+2.cos\left(a-b\right)=2+2.cos\frac{\pi}{3}=3\)

\(B=cos^2a+sin^2b+2cosa.sinb+cos^2b+sin^2a-2sina.cosb\)

\(=2-2\left(sina.cosb-cosa.sinb\right)\)

\(=2-2sin\left(a-b\right)=2-2sin\frac{\pi}{3}=2-\sqrt{3}\)

a) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).

b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .

c) \(2\left(sin^6\alpha+cos^6\alpha\right)+1=2\left(sin^2\alpha+cos^2\alpha\right)\)\(\left(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha\right)+1\)
\(=2\left(sin^4\alpha+cos^4\alpha-sin^2\alpha cos^2\alpha\right)+1\)
\(=2\left(sin^4\alpha+cos^4\alpha\right)+sin^2\alpha-sin^2\alpha cos^2\alpha+\)\(cos^2\alpha-sin^2\alpha cos^2\alpha\)
\(=2\left(sin^4\alpha+cos^4\alpha\right)+sin^2\alpha\left(1-cos^2\alpha\right)+\)\(cos^2\alpha\left(1-sin^2\alpha\right)\)
\(=2\left(sin^4\alpha+cos^4\alpha\right)+sin^2\alpha.sin^2\alpha+cos^2\alpha.cos^2\alpha\)
\(=3\left(sin^4\alpha+cos^4\alpha\right)\).