a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)

\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)

\(=\dfrac{5}{2}\sqrt{5}+7\)

b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)

\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)

\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)

\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)

\(=\dfrac{1}{\sin25^0}-1\)

\(=\dfrac{1-\sin25^0}{\sin25^0}\)

\(A=sin^225+cos^2\left(90-65\right)-tan35+tan\left(90-55\right)-\frac{cot32}{cot\left(90-58\right)}\)

\(=sin^225+cos^225-tan35+tan35-\frac{cot32}{cot32}\)

\(=1-0-1=0\)

?

a/ \(\tan40.\cot40+\frac{\sin50}{\cos40}\)

\(=1+\frac{\cos40}{\cos40}=1+1=2\)

Đề yêu cầu làm gì bạn

giúp mình đi!!

\(\sin^225^o+\sin^265^o-\tan35^o+\cot55^o-\frac{\cot32^o}{tan58^o}\)

\(=\cos^265^o+\sin^265^o-\cot55^{^{ }o}+\cot55^o-\frac{\tan58^o}{\tan58^o}\)

\(=1-0-1\)

\(=0\)

nhớ k cho mik nha ^^