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Em thử thôi chứ ko chắc đâu:((
Xét dạng tổng quát \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
Suy ra \(A=\frac{1}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{100-\sqrt{100}}{100}\)
Ta có :
\(A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+.....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}+........+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\) \(=1-\dfrac{1}{\sqrt{100}}< 1\)
Vậy \(A< 1\)
Ta có công thức tổng quát
\(\dfrac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)Vậy \(P=\dfrac{1}{\sqrt{2}.1+\sqrt{1}.2}+\dfrac{1}{\sqrt{3}.2+\sqrt{2}.3}+...+\dfrac{1}{\sqrt{100}.99+\sqrt{99}.100}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Ta có công thức tổng quát
\(\dfrac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)Vậy \(P=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+...+\dfrac{1}{99\sqrt{100}+100\sqrt{99}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-1}{\left(1+\sqrt{2}\right).\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right).\left(\sqrt{3}-2\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-1=10-1=9\)
Ta có:\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}^2-\sqrt{n}^2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Do đó:
\(\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1;\dfrac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2};...;\dfrac{1}{\sqrt{99}-\sqrt{100}}=\sqrt{100}-\sqrt{99}\)
Đến đây bạn tự giải tiếp nhé.