Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
C1:A2=\(3-\sqrt{5}+3+\sqrt{5}+2\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
A2=\(6+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
A2=\(6+2\sqrt{9-5}\)
A2=6+4=10
A=\(\sqrt{10}\)
\(A=\sqrt{\left(3+\sqrt{5}\right)}+\sqrt{\left(3-\sqrt{5}\right)}\)
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right).\left(3-\sqrt{5}\right)}\)
\(A^2=6+2\sqrt{9-5}\)
\(A^2=6+2\sqrt{4}\)
\(A^2=8\)
\(\Rightarrow A=\sqrt{8}\)
\(\left(3+\frac{\sqrt{5}}{\sqrt{10}}+\sqrt{3}+\sqrt{5}\right)-\left(3-\frac{\sqrt{5}}{\sqrt{10}}+\sqrt{3}-\sqrt{5}\right)=\sqrt{34.64911064}\)
\(tana=\sqrt{3}\)
=>\(\dfrac{sina}{cosa}=\sqrt{3}\)
=>\(sina=\sqrt{3}\cdot cosa\)
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+3=4\)
=>\(cos^2a=\dfrac{1}{4}\)
=>\(cosa=\dfrac{1}{2}\)
=>\(sina=\dfrac{\sqrt{3}}{2}\)
\(A=\dfrac{sin^2a-cos^2a}{sina\cdot cosa}\)
\(=\dfrac{\dfrac{3}{4}-\dfrac{1}{4}}{\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}}=\dfrac{2}{4}:\dfrac{\sqrt{3}}{4}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\)
áp dụng công thức sin2a+cos2a=1
A= sin2a +cos2a-2sina.cosa-sin2a-cos2a+2sina.cosa = 0
B=(sỉn2a+cos2a)2 =12 =1
C= cos2a(cos2a+sin2a)+ sin2a=cos2a+sin2a=1
D=sin2a(sin2p+cos2p)+cos2a=sin2a+cos2a=1
E= (sin2a+cos2a)(sin4a-sin2a.cos2a+cos4a)+3sin2a.cos2a
=sin4a+2sin2a.cos2a+ cos4a=(sin2a+cos2a)2=1
\(270^0< a< 360^0\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{\sqrt{5}}{3}\)