cos20,sin65,cos28,sin40,cos88 

Giải thích các bước giải:

 đổi sin40=cos(90-40)=cos50

sin65=cos(90-65)=cos25

\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)

\(\frac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

\(\frac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)

\(\frac{\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|}{\sqrt{2}}\)

\(\frac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}\)

\(=\sqrt{2}\)

Đặt \(2n+1=a^2,3n+1=b^2\).

\(15n+8=9\left(2n+1\right)-\left(3n+1\right)=9a^2-b^2=\left(3a-b\right)\left(3a+b\right)\)

Hiển nhiên \(3a+b>1\).

Nếu \(3a-b=1\Rightarrow b+1⋮3\).

mà \(b^2\equiv1\left(mod3\right)\Leftrightarrow b\equiv1\left(mod3\right)\Leftrightarrow b\equiv2\left(mod3\right)\)mâu thuẫn

do đó \(3a-b\ne1\).

Do đó \(15n+8\)là hợp số. 

\(\sqrt{96}.\sqrt{125}\)

\(\sqrt{16.6}\sqrt{25.5}\)

\(4.5\sqrt{6.5}\)

\(20\sqrt{30}\)

\(b,\sqrt{a^4b^5}\)

\(a^2b^2\sqrt{b}\)

\(c,\sqrt{a^6b^{11}}\)

\(a^3b^5\sqrt{b}\)

\(d,\sqrt{a^3\left(1-a\right)^4}\)

\(a\left(1-a\right)^2\sqrt{a}\)

\(x\sqrt{x}+y\sqrt{y}+x-y\)

\(=\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y+\sqrt{x}-\sqrt{y}\right)\)

#H

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