a,Để \(\sqrt{x^2-8x-9}\) có nghĩ thì

 \(x^2-8x-9\ge0\)

\(\Leftrightarrow x^2+x-9x-9\ge0\)

\(\Leftrightarrow x\left(x+1\right)-9\left(x+1\right)\ge0\)

\(\Leftrightarrow\left(x+1\right)\left(x-9\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1\ge0\\x-9\ge0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge-1\\x\ge9\end{cases}\Rightarrow}x\ge9\)

\(or\orbr{\begin{cases}x+1\le0\\x-9\le0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\le-1\\x\le9\end{cases}\Rightarrow}x\le-1\)

\(Để\sqrt{4-9x^2}\text{có nghĩa}\)

\(\Rightarrow4-9x^2\ge0\)

\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)\ge0\)

\(\Leftrightarrow-\frac{2}{3}\le x\le\frac{2}{3}\)

x² - 8x - 9  ≥ 0

<=> (x+1)(x-9)\(\ge\)0

<=> \(\hept{\begin{cases}x+1\ge0\\x-9\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-9\le0\end{cases}}\)

<=> \(\orbr{\begin{cases}x\ge9\\x\le-1\end{cases}}\)

\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)

\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)

\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)

\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)

\(=14\)

\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)

\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)

\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)

\(=1+\sqrt{2}\)

a) Ta có: \(\sin^2a^o=\cos^2\left(90^o-a^o\right)\)

Biểu thức trên

\(=\left(\sin^21^o+\sin^o89\right)+\left(\sin^22^o+\sin^288^o\right)+...+\left(\sin^244^o+\sin^246^o\right)+\sin^245^o\)

\(=\left(\sin^21^o+\cos^21^o\right)+\left(\sin^22^o+\cos^22^o\right)+...+\left(\sin^244^o+\cos^246^o\right)+\sin^245^o\)

\(=1+1+..+1+\sin^245^o=44+\frac{1}{2}=\frac{89}{2}\)

b) 

Ta có: \(\sin^2x+\cos^2x=1\)

\(0^o< x< 90^o\)

=> \(0< \sin x;\cos x< 1\)

Ta có:  \(\frac{\sin^2x+\cos^2x}{\text{​​}\text{​​}\sin x.\cos x}=\frac{1}{\frac{12}{25}}=\frac{25}{12}\Leftrightarrow\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{25}{12}\)

\(\Leftrightarrow\tan x+\frac{1}{\tan x}=\frac{25}{12}\Leftrightarrow\tan^2x-\frac{25}{12}\tan x+1=0\)

Đặt t =tan x => có phương trình bậc 2 ẩn t => Giải đen ta => ra đc t => ra đc tan t

\(\Leftrightarrow\orbr{\begin{cases}\tan x=\frac{3}{4}\\\tan x=\frac{4}{3}\end{cases}}\)