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Ta có :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
Ta có:
\(\frac{1}{\sqrt{x}+\sqrt{x-1}}=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}=\sqrt{x}-\sqrt{x-1}\)
Do đó:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
=\(\frac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\)+\(\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\sqrt{2}-\sqrt{3}}\)+.....+\(\frac{\sqrt{99}-\sqrt{100}}{\left(\sqrt{99}+\sqrt{100}\right).\left(\sqrt{99}-\sqrt{100}\right)}\)
=\(\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+....+\sqrt{100}-\sqrt{99}\)
=\(-1+\sqrt{100}\)
=9
=1.1.3.3.5.5...99.99/1.3.3.5.5.7.....99.101
=(1.3.5..99/1.3.5....99).(1.3.5....99/3.5.7...101)
=1.1/101
=1/101
=1.1.3.3.5.5...99.99/1.3.3.5.5.7.....99.101
=(1.3.5..99/1.3.5....99).(1.3.5....99/3.5.7...101)
=1.1/101
=1/101
......................?
mik ko biết
mong bn thông cảm
nha ................
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{100}-1-\frac{1}{2}-...-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)