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

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

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

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

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

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)

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

Ta có: \(A^2=\left[\left(\sqrt{9-\sqrt{17}}\right).\left(\sqrt{9+\sqrt{17}}\right)\right]=\left(9-\sqrt{17}\right).\left(9+\sqrt{17}\right)\)

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

\(=>A=\sqrt{64}=8\)
 

Xét vế trái:

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

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

\(=\left|\sqrt{\frac{17}{2}}-\sqrt{\frac{1}{2}}\right|.\left|\sqrt{\frac{17}{2}}+\sqrt{\frac{1}{2}}\right|\)

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

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

\(=\frac{17}{2}-\frac{1}{2}=8\)

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

(Nhớ k cho mình với nha!)

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 )

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

a) Sai đề 

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

Vậy VT=VP

\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{9^2-17}=\sqrt{64}=8\)

\(2\sqrt{2}\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\) \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{2-2\sqrt{10}+5}+\sqrt{2}=\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{2}.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\left|\sqrt{5}-\sqrt{2}\right|+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\) \(\sqrt{\sqrt{3}+\sqrt{2}}.\sqrt{\sqrt{3}-\sqrt{2}}=\sqrt{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}=\sqrt{3-2}=\sqrt{1}=1\) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\left[\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4-\sqrt{15}}\right)\right]=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right);\left[\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\right]^2=2.\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\Rightarrow\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4}=2\left(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}>0\right)\)

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) \(9+4\sqrt{5}=4+4\sqrt{5}+5=2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{5}+2\right)^2\left(ĐPCM\right)\)

a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\left(đpcm\right)\)

b)\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\left(đpcm\right)\)

c)\(\left(4-\sqrt{7}\right)^2=16-8\sqrt{7}+7=23-8\sqrt{7}\left(đpcm\right)\)

d)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}=4+\sqrt{7}-\sqrt{7}=4\left(đpcm\right)\)