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25 tháng 2 2017

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)

25 tháng 2 2017

vì A nhỏ hơn hoặc bằng 3 và B nhỏ hơn hoặc bằng 5 =>A+B nhỏ hơn hoặc bằng 8 => A+B<10

14 tháng 10 2017

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: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

7 tháng 8 2017

~ ~ ~

\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\)

\(=\sqrt{\dfrac{37}{4}-\sqrt{\left(3\sqrt{5}+2\right)^2}}\)

\(=\sqrt{\dfrac{29}{4}-3\sqrt{5}}\)

\(=\sqrt{\dfrac{29-12\sqrt{5}}{4}}\)

\(=\sqrt{\dfrac{\left(2\sqrt{5}-3\right)^2}{4}}\)

\(=\dfrac{\sqrt{5}}{2}-\dfrac{3}{4}\)

\(=\dfrac{1}{2}\left(\sqrt{5}-\dfrac{3}{2}\right)\)

\(>\sqrt{5}-\dfrac{3}{2}=B\)

~ ~ ~

\(C=\dfrac{16\sqrt{36}-20\sqrt{48}+10\sqrt{3}}{\sqrt{12}}\)

\(=\dfrac{96-80\sqrt{3}+10\sqrt{3}}{\sqrt{12}}\)

\(=\dfrac{96-70\sqrt{3}}{2\sqrt{3}}\)

\(=16\sqrt{3}-35\)

\(>16\sqrt{3}-36=B\)

~ ~ ~

8 tháng 8 2017

Cau A sao sao ak ban oi

12 tháng 8 2023

a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)

\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)

\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)

\(=\left(10-4+6\right)\sqrt{2}\)

\(=12\sqrt{2}\)

b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)

\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)

\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)

\(=\left(8-15+15-3\right)\sqrt{5}\)

\(=5\sqrt{5}\)

c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)

\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)

\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)