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a) \(\sqrt{121}=11\)
\(\sqrt{12321}=111\)
\(\sqrt{1234321}=1111\)
b) \(\sqrt{123454321}=11111\)
\(\sqrt{12345654321}=111111\)
\(\sqrt{1234567654321}=1111111\)
a) \(\sqrt{1}=1\)
\(\sqrt{1+2+1}=2\)
\(\sqrt{1+2+3+2+1}=3\)
b) \(\sqrt{1+2+3+4+3+2+1}=4\)
\(\sqrt{1+2+3+4+5+4+3+2+1}=5\)
\(\sqrt{1+2+3+4+5+6+5+4+3+2+1}=6\)
1 + 2 + 3 + 4 + 3 + 2 + 1 = 4 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 6
Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.
123454321 = 11111 12345654321 = 111111 1234567654321 = 1111111