\(\sqrt{24}< \sqrt{25}=5;\sqrt{45}< \sqrt{49}=7\)

\(->\sqrt{24}+\sqrt{45}< 5+7=12\)

A B C

Sin B = \(\frac{AC}{BC}\); cos B = \(\frac{AB}{BC}\) ; tgB = \(\frac{AC}{AB}\); cot gB = \(\frac{AB}{BC}\)

Do góc B và C là hai góc phụ nhau nên : 

sin C = cos B = \(\frac{AB}{BC};cosB=\frac{AB}{BC};cosC=sinB=\frac{AC}{BC}\)

\(tgC=cotgB=\frac{AB}{BC};cotgC=tgB=\frac{AC}{AB}\)

Chúc bạn học tốt !!!

DÙNG GÓC NHỌN 2 .BIẾT RẰNG TAN 2=\(\frac{4}{5}\)

a.\(DKXD:x\ge1\)

b.\(A=\sqrt{x-\sqrt{x^2-4x+4}}=\sqrt{x-\sqrt{\left(x-2\right)^2}}=\sqrt{x-|x-2|}=\orbr{\begin{cases}\sqrt{2}\left(x\ge2\right)\\2x-2\left(1\le x< 2\right)\end{cases}}\)

Ap dung BDT Bun-hia-cop-xki ta co:

\(\left(a.\cos\beta+b.\sin\beta\right)^2\le\left(a^2+b^2\right)\left(\cos^2\beta+\sin^2\beta\right)=a^2+b^2\)

\(\Rightarrow-\sqrt{a^2+b^2}\le a.\cos\beta+b.\sin\beta\le\sqrt{a^2+b^2}\)

Dau '=' xay ra khi \(\frac{a}{\cos\beta}=\frac{b}{\sin\beta}\)

\(A=\frac{1}{11.m.n}.m.n.\sqrt{\frac{121.m^2}{n^6}}=\frac{1}{11}.\frac{11.m}{n^3}=\frac{m}{n^3}\)

\(B=2\left(m+n\right).\sqrt{\frac{1}{m^2+2mn+n^2}}=2\left(m+n\right).\sqrt{\frac{1}{\left(m+n\right)^2}}=2\)