a) Ta có: \(9+4\sqrt{5}\)

\(=5+2\cdot\sqrt{5}\cdot2+4\)

\(=\left(\sqrt{5}+2\right)^2\)(đpcm)

b) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

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

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

=-2(ddpcm)

c) Ta có: \(\left(4-\sqrt{7}\right)^2\)

\(=16-2\cdot4\cdot\sqrt{7}+7\)

\(=23-8\sqrt{7}\)(đpcm)

d) Ta có: \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}\)

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

\(=3-2\sqrt{2}+2\sqrt{2}=3\)(đpcm)

\(a.VT=4+4\sqrt{5}+5=2^2+4\sqrt{5}+\sqrt{5}^2=\left(2+\sqrt{5}\right)^2=VP\)

\(b.\) Dựa vào câu a ta có: \(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

\(VT=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2=VP\)

\(c.VT=16-8\sqrt{7}+7=4^2-8\sqrt{7}+\sqrt{7}^2=\left(4-\sqrt{7}\right)^2=VP\)

\(d.\) 

Ta có: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)

\(VT=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3=VP\)

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)

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

= \(\sqrt{9^2-\sqrt{17}^2}\)

= \(\sqrt{81-17}\)

= \(\sqrt{64}\)

= 8

b/ =\(2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)

= 9

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

b: \(=\sqrt{81-17}=8\)

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

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