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\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{\left(xy+yz+zx\right)^2}{x^2y^2z^2}\)(1) với x+y+z=0. Bạn quy đồng vế trái (1) dc \(\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}=\frac{\left(xy+yz+zx\right)^2-2\left(x+y+z\right)xyz}{x^2y^2z^2}\)
1/ a/ \(\sqrt{\left(6+2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^6}-\sqrt{\left(\sqrt{5}-1\right)^6}\)
\(=\left(\sqrt{5}+1\right)^3-\left(\sqrt{5}-1\right)^3\)
\(=32\)
b/ \(\sqrt{\left(3-2\sqrt{2}\right)\left(4-2\sqrt{3}\right)}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{2}-1\right)\left(\sqrt{3}-1\right)\)
\(=\sqrt{6}-\sqrt{2}-\sqrt{3}+1\)
Câu 3/ \(A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{2}}}}}\)
\(< \sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{4}}}}}=2\)
Ta lại có:
\(A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{2}}}}}>\sqrt{2}>1\)
\(\Rightarrow1< A< 2\)
Vậy \(A\notin N\)
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3.\left(2-1\right)}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5\left(3-2\right)}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011\left(2006-2005\right)}\)
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011}\)
Nhận xét: (a-b)2 \(\ge\) 0 => a2 + b2 \(\ge\) 2ab
Áp dụng ta có: \(3=\left(\sqrt{2}\right)^2+\left(\sqrt{1}\right)^2\ge2.\sqrt{2}.\sqrt{1}\)
\(5=\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2\ge2.\sqrt{3}.\sqrt{2}\)
...
\(4011=\left(\sqrt{2006}\right)^2+\left(\sqrt{2005}\right)^2\ge2.\sqrt{2006}.\sqrt{2005}\)
=> \(VT<\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{2.\sqrt{2}.\sqrt{1}}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{2.\sqrt{3}.\sqrt{2}}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{2.\sqrt{2006}.\sqrt{2005}}\)
=> \(VT<1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}...+\frac{1}{\sqrt{2005}}-\frac{1}{\sqrt{2006}}=1-\frac{1}{\sqrt{2006}}\)
=> điều phải chứng minh
\(\sqrt{2+\sqrt{3}}=\sqrt{\frac{1}{2}\left(4+2\sqrt{3}\right)}=\sqrt{\frac{1}{2}}\sqrt{3+2\sqrt{3}+1}=\sqrt{\frac{1}{2}}\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{\frac{1}{2}}.\left(\sqrt{3}+1\right)=\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\left(đpcm\right)\)