Em nào thích làm thì làm nhé
Chứng minh A > 1
\(A=\frac{1}{1\sqrt{2}+2\sqrt{1}}\) \(+\frac{1}{2\sqrt{3}+3\sqrt{2}}\)\(+\frac{1}{3\sqrt{4}+4\sqrt{3}}\)\(+\frac{1}{400\sqrt{401}+401\sqrt{400}}\)
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17 tháng 10 2018
CM bđt phụ nhá: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) \(\left(n\inℕ^∗\right)\)
\(VT=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right)\left(\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right)}\)
\(VT=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(VT=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n^2+2n+1-n^2-n\right)}\)
\(VT=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
\(VT=\frac{\left(n+1\right)\sqrt{n}}{n\left(n+1\right)}-\frac{n\sqrt{n+1}}{n\left(n+1\right)}\)
\(VT=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=VP\)
Áp dụng vào A ta có :
\(A=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{399\sqrt{400}+400\sqrt{399}}\) ( olm bị lỗi nên ko dám viết nhìu )
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(A=1-\frac{1}{20}=\frac{19}{20}\)
Vậy \(A=\frac{19}{20}\)
Chúc bạn học tốt ~
NH
16 tháng 9 2020
a)\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2.n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b)\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\( S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
16 tháng 9 2020
\(a,\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(n+1-n\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{\sqrt{n-1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
27 tháng 10 2020
Xét phân thức phụ sau:
Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\frac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thay vào ta được:
\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
NN
27 tháng 10 2020
Đặt biểu thức đã cho là A
Tổng quát ta có: Với \(a\inℕ^∗\)ta có:
\(\frac{1}{\left(a+1\right)\sqrt{a}+a.\sqrt{a+1}}=\frac{\left(a+1\right)-a}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}\)
\(=\frac{\left(\sqrt{a+1}-\sqrt{a}\right)\left(\sqrt{a+1}+\sqrt{a}\right)}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}\)
\(=\frac{\sqrt{a+1}}{\sqrt{a}.\sqrt{a+1}}-\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}\)
Áp dụng kết quả trên ta có:
Với \(n=1\)\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Với \(n=2\)\(\Rightarrow\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
Với \(n=3\)\(\Rightarrow\frac{1}{4\sqrt{3}+3\sqrt{4}}=\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\)
.....................
Với \(n=399\)\(\Rightarrow\frac{1}{400\sqrt{399}+399\sqrt{400}}=\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+......+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
ML
23 tháng 5 2018
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 \(a=b=2\)
TP
3 tháng 8 2017
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:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
Ta có:
\(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}.\text{ Vì thế, }A=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...-\frac{1}{\sqrt{401}}< 1.\)
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