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Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
\(=\dfrac{\left(1+\dfrac{99}{2}+1+\dfrac{98}{3}+...+1+\dfrac{1}{100}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
=101-2
=99
Ta có:
\(\frac{1}{n\sqrt{\left(n+1\right)}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{\left(n+1\right)}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào ta được
\(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{99\sqrt{100}+100\sqrt{99}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
1.Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1 .Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200