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

b/ \(\left(\sqrt{2}-1\right)^2=2-2\sqrt{2}+1=\sqrt{9}-\sqrt{8}\)

a)  Bình phương vế trái, ta được:

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

\(\Leftrightarrow\left(9-\sqrt{17}\right).\left(9+\sqrt{17}\right)\)

\(\Leftrightarrow81-17=64=8^2\)

\(\Rightarrow\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8\left(đpcm\right)\)

b) Ta có: \(\left(\sqrt{2}-1\right)^2=\left(\sqrt{2}\right)^2-2\sqrt{2}+1=2-2\sqrt{2}+1=3-2\sqrt{2}=\sqrt{9}-\sqrt{8}\) (đpcm)

\(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{2^2+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\sqrt{\left(2+\sqrt{3}\right)^2}\)

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

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

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

a) \(VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)

=\(\sqrt{9^2-\left(\sqrt{17}\right)^2}=\sqrt{81-17}=\sqrt{64}=8=VP\)

b) \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

=\(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}=9=VP\)

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 )

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

\(\sqrt{9-\sqrt[]{17}}+\sqrt{9+\sqrt{7}=8}\)

\(9-17+9+7=8\)

\(-8+16=8\)

Sửa đề: \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8\)

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

a) Sai đề 

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

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

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

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

=8-8

=0

Bài 2:

a)

\(\sqrt{9-\sqrt{17}}-\sqrt{9+\sqrt{17}}=\sqrt{\frac{18-2\sqrt{17}}{2}}-\sqrt{\frac{18+2\sqrt{17}}{2}}\)

\(=\sqrt{\frac{17+1-2\sqrt{17}}{2}}-\sqrt{\frac{17+1+2\sqrt{17}}{2}}=\sqrt{\frac{(\sqrt{17}-1)^2}{2}}-\sqrt{\frac{(\sqrt{17}+1)^2}{2}}\)

\(=\frac{\sqrt{17}-1}{\sqrt{2}}-\frac{\sqrt{17}+1}{\sqrt{2}}=-\sqrt{2}\)

b)

\(2\sqrt{2}(\sqrt{3}-2)+(1+2\sqrt{2})^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+(1+4\sqrt{2}+8)-2\sqrt{6}\)

\(=1+8=9\)

Bài 1:

a)

\(\frac{\sqrt{6}+\sqrt{16}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+4}{2(\sqrt{3}+\sqrt{7})}=\frac{1}{2}.\frac{(\sqrt{6}+4)(\sqrt{7}-\sqrt{3})}{(\sqrt{3}+\sqrt{7})(\sqrt{7}-\sqrt{3})}\)

\(=\frac{(4+\sqrt{6})(\sqrt{7}-\sqrt{3})}{8}\)

b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{16}-\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+1)(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)

a) Ta có: \(VT=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)

\(=\sqrt{\left(9-\sqrt{17}\right)\cdot\left(9+\sqrt{17}\right)}\)

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

b) Ta có: \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)

=9=VP(đpcm)