chứng minh rằng
\(\sqrt{9-\sqrt{17}}\) . \(\sqrt{9+\sqrt{17}}\) = 8
\(2\sqrt{2}\)\(\left(\sqrt{3}-2\right)\) + \(\left(1+2\sqrt{2}\right)^2\)- \(2\sqrt{6}\) = 9
\(\sqrt{7-2\sqrt{10}}\) + \(\sqrt{2}\) = \(\sqrt{5}\)
\(\sqrt{\sqrt{3}+\sqrt{2}}\) . \(\sqrt{\sqrt{3}-\sqrt{2}}\) = 1
\(\left(4+\sqrt{15}\right)\) \(\left(\sqrt{10}-\sqrt{6}\right)\) \(\sqrt{4-\sqrt{15}}\) = 2
\(\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}=\sqrt{81-17}=\sqrt{64}=8\)
Vậy VT=VP
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{9^2-17}=\sqrt{64}=8\)
\(2\sqrt{2}\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\) \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{2-2\sqrt{10}+5}+\sqrt{2}=\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{2}.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\left|\sqrt{5}-\sqrt{2}\right|+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\) \(\sqrt{\sqrt{3}+\sqrt{2}}.\sqrt{\sqrt{3}-\sqrt{2}}=\sqrt{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}=\sqrt{3-2}=\sqrt{1}=1\) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\left[\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4-\sqrt{15}}\right)\right]=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right);\left[\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\right]^2=2.\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\Rightarrow\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4}=2\left(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}>0\right)\)
