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a) \(\sqrt{1}=1\)
\(\sqrt{1+2+1}=2\)
\(\sqrt{1+2+3+2+1}=3\)
b) \(\sqrt{1+2+3+4+3+2+1}=4\)
\(\sqrt{1+2+3+4+5+4+3+2+1}=5\)
\(\sqrt{1+2+3+4+5+6+5+4+3+2+1}=6\)
a) \(\sqrt{121}=11\)
\(\sqrt{12321}=111\)
\(\sqrt{1234321}=1111\)
b) \(\sqrt{123454321}=11111\)
\(\sqrt{12345654321}=111111\)
\(\sqrt{1234567654321}=1111111\)
\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)
\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)
a
.\(\sqrt{1}=1\)
\(\sqrt{1+2+1}=\sqrt{4}=2\)
\(\sqrt{1+2+3+2+1}=\sqrt{9}=3\)
b,
\(\sqrt{1+2+3+4+3+2+1}=\sqrt{16}=4\)
\(\sqrt{1+2+3+4+5+4+3+2+1}=\sqrt{25}=5\)
\(\sqrt{1+2+3+4+5+6+5+4+3+2+1}=\sqrt{36}=6\)
*Nhận xét:
+\(\sqrt{1+...+10+...1}=10\)
+\(\sqrt{1+2+...+100+1}=100\)
+\(\sqrt{1+2+...n+...1}=n;n\in N\)*