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Ta có:
\(A=\sqrt{5\sqrt{5\sqrt{5...\sqrt{5}}}}< \sqrt{5\sqrt{5\sqrt{5...\sqrt{25}}}}=5\)
\(B=\sqrt{20+\sqrt{20+...+\sqrt{20}}}< \sqrt{20+\sqrt{20+...+\sqrt{25}}}=5\)
\(\Rightarrow A+B< 5+5=10\)
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
~ ~ ~
\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\)
\(=\sqrt{\dfrac{37}{4}-\sqrt{\left(3\sqrt{5}+2\right)^2}}\)
\(=\sqrt{\dfrac{29}{4}-3\sqrt{5}}\)
\(=\sqrt{\dfrac{29-12\sqrt{5}}{4}}\)
\(=\sqrt{\dfrac{\left(2\sqrt{5}-3\right)^2}{4}}\)
\(=\dfrac{\sqrt{5}}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}\left(\sqrt{5}-\dfrac{3}{2}\right)\)
\(>\sqrt{5}-\dfrac{3}{2}=B\)
~ ~ ~
\(C=\dfrac{16\sqrt{36}-20\sqrt{48}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-80\sqrt{3}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-70\sqrt{3}}{2\sqrt{3}}\)
\(=16\sqrt{3}-35\)
\(>16\sqrt{3}-36=B\)
~ ~ ~
\(A=\sqrt{47+\sqrt{5}}\cdot\sqrt{47-\sqrt{5}}\)
\(=\sqrt{2204}=2\sqrt{551}\)
\(B=5-2\sqrt{6}+10+\sqrt{6}=15-\sqrt{6}\)
Ta có: \(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5}}}}< \sqrt{5\sqrt{5\sqrt{5...\sqrt{25}}}}=...=5\)
\(\sqrt{20+\sqrt{20+...+\sqrt{20}}}< \sqrt{20+\sqrt{20+...+\sqrt{25}}}=...=5\)
Vậy A+B<5+5=10 (ĐPCM)
vì A nhỏ hơn hoặc bằng 3 và B nhỏ hơn hoặc bằng 5 =>A+B nhỏ hơn hoặc bằng 8 => A+B<10