a) \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=\sqrt{64}=8\)

b)\(\sqrt{9\left(3-a\right)^2}=3\left|3-a\right|=3\left(a-3\right)\)(vì a > 3)

\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)

\(=\sqrt{\left(\sqrt{9}\right)^2}-\sqrt{\left(\sqrt{17}\right)^2}\)

\(\sqrt{9\left(3-a\right)^2}\)

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

\(=3\left(3-a\right)\)

\(=3-3a\)

\(2+\sqrt{17-4.\sqrt{9+4.\sqrt{5}}}\)

\(=2+\sqrt{17-4.\sqrt{4+4.\sqrt{5}+5}}\)

\(=2+\sqrt{17-4.\sqrt{2^2+2.2.\sqrt{5}+\sqrt{5}^2}}\)

\(=2+\sqrt{17-4.\sqrt{\left(2+\sqrt{5}\right)^2}}\)

\(=2+\sqrt{17-4.\left(2+\sqrt{5}\right)}\)

\(=2+\sqrt{17-8+4.\sqrt{5}}\)

\(=2+\sqrt{9+4.\sqrt{5}}\)

\(=2+\sqrt{4+4.\sqrt{5}+5}\)

\(=2+\sqrt{2^2+2.2.\sqrt{5}+\sqrt{5}^2}\)

\(=2+\sqrt{\left(2+\sqrt{5}\right)^2}\)

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

'\(=4+\sqrt{5}\)

a.

\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\\ =\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\\ =\sqrt{81-17}\\ =\sqrt{64}\\=8\)

\(a.VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=8=VP\)

\(b.\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=3\sqrt{3}-\sqrt{2}\) ( thiếu đề )

\(VT=\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=\dfrac{1}{3-2\sqrt{3}.\sqrt{2}+2}+\dfrac{2}{3+2\sqrt{3}.\sqrt{2}+2}=\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-2\sqrt{2}=3\sqrt{3}-\sqrt{2}=VP\)

Không hiểu sao cứ gửi ảnh nó lại bị lộn xộn nên bạn cố nhìn nhé

( ͡°( ͡° ͜ʖ( ͡° ͜ʖ ͡°)ʖ ͡°) ͡°)

a: \(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\)

b: \(=\sqrt{81-17}=8\)

\(\sqrt{17-\sqrt{33}}\cdot\sqrt{17+\sqrt{33}}\)

\(=\sqrt{\left(17-\sqrt{33}\right)\left(17+\sqrt{33}\right)}\)

\(=\sqrt{17^2-\left(\sqrt{33}\right)^2}\)

\(=\sqrt{289-33}\)

\(=\sqrt{256}\)

\(=\sqrt{16^2}\)

\(=16\)

Đk:\(-\sqrt{17}\le x\le\sqrt{17}\)

Đặt \(t=x+\sqrt{17-x^2}\left(t>0\right)\)

\(\Rightarrow t^2=17+2x\sqrt{17-x^2}\)

\(\Rightarrow x\sqrt{17-x^2}=\frac{t^2-17}{2}\)

thay vào pt 

\(t+\frac{t^2-17}{2}=9\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-7\left(loai\right)\\t=5\left(tm\right)\end{array}\right.\)

\(\Rightarrow x+\sqrt{17-x^2}=5\)

\(\Leftrightarrow\sqrt{17-x^2}=5-x\)

Với \(x< \sqrt{17}\) bình 2 vế ta có:

\(17-x^2=x^2-10x+25\)

\(\Leftrightarrow2x^2-10x+8=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=4\end{cases}\left(tm\right)}\)

dòng cuối là \(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=4\end{array}\right.\)(thỏa mãn)

\(K=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{81-17}-8=\sqrt{64}-8=8-8=0\)

\(=\sqrt{81-17}-8\)

=8-8

=0