Chứng minh rằng: \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+\(\frac{1}{\sqrt{3}}\)+......+\(\frac{1}{\sqrt{121}}\)>11
Những câu hỏi liên quan
cho \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{120}+\sqrt{121}}\)
\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{35}}\)
Chứng minh rằng A<B
(Bình Định)
Cho \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{120}+\sqrt{121}}\)
và \(B=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{35}}\).
Chứng minh rằng \(B>A\).
Cho A = \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{120}+\sqrt{121}}\)
B = \(1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{35}}\)
Chưnhs minh rằng: B > A
\(A=\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{121}-\sqrt{120}}{\left(\sqrt{121}-\sqrt{120}\right)\left(\sqrt{121}+\sqrt{120}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{121}-\sqrt{120}\)
\(=\sqrt{121}-1=10\)
\(B=\frac{2}{2.\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{35}}\)
\(B>\frac{1}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)
\(B>2\left(\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+...+\frac{\sqrt{36}-\sqrt{35}}{\left(\sqrt{36}-\sqrt{35}\right)\left(\sqrt{36}+\sqrt{35}\right)}\right)\)
\(B>2\left(\sqrt{36}-\sqrt{1}\right)=10\Rightarrow B>A\)
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\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{120}+\sqrt{121}}\)
\(B=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{35}}\)
Chứng minh B>A
cho A= \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{120}+\sqrt{121}}\); B = \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{35}}\)
Chứng minh A<B
Chứng minh rằng \(\frac{1}{3\sqrt[2]{2}}+\frac{1}{4\sqrt[3]{3}}+\frac{1}{5\sqrt[3]{4}}+.....+\frac{1}{1000\sqrt[3]{999}}< \frac{11}{5}\)
Tính giá trị biểu thức:text{a) }frac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+frac{1}{sqrt{4}+sqrt{5}}+...+frac{1}{sqrt{2010}+sqrt{2011}}text{b) }frac{1}{2sqrt{1}+1sqrt{2}}+frac{1}{3sqrt{2}+2sqrt{3}}+frac{1}{4sqrt{3}+3sqrt{4}}+...+frac{1}{121sqrt{120}+120sqrt{121}}text{c) }sqrt{1+frac{1}{1^2}+frac{1}{2^2}}+sqrt{1+frac{1}{2^2}+frac{1}{3^2}}+sqrt{1+frac{1}{3^2}+frac{1}{4^2}}+...sqrt{+1+frac{1}{2010^2}+frac{1}{2011^2}}
Đọc tiếp
Tính giá trị biểu thức:
\(\text{a) }\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{2010}+\sqrt{2011}}\)
\(\text{b) }\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{121\sqrt{120}+120\sqrt{121}}\)
\(\text{c) }\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...\sqrt{+1+\frac{1}{2010^2}+\frac{1}{2011^2}}\)
Bài 1. So sánha) sqrt{2009}-sqrt{2008}và sqrt{2007}-sqrt{2006}b) sqrt{11+sqrt{96}}và frac{2sqrt{2}}{1+sqrt{2}-sqrt{3}}Bài 2. Tính tổngTfrac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...+frac{1}{sqrt{2007}-sqrt{2008}}Dfrac{1}{2sqrt{1}+1sqrt{2}}+frac{1}{3sqrt{2}+2sqrt{3}}+frac{1}{4sqrt{3}+3sqrt{4}}+...+frac{1}{120sqrt{121}+121sqrt{120}}ai cứu mk ikk
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Bài 1. So sánh
a) \(\sqrt{2009}-\sqrt{2008}\)và \(\sqrt{2007}-\sqrt{2006}\)
b) \(\sqrt{11+\sqrt{96}}\)và \(\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)
Bài 2. Tính tổng
\(T=\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...+\frac{1}{\sqrt{2007}-\sqrt{2008}}\)
\(D=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{120\sqrt{121}+121\sqrt{120}}\)
ai cứu mk ikk
Bài 2:
\(D=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{120\sqrt{121}+121\sqrt{120}}\)
Với mọi \(n\inℕ^∗\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}]^2-\left(n\sqrt{n+1}\right)^2}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\left(\sqrt{n}+1\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}}{n\left(n+1\right)}-\frac{n\sqrt{n+1}}{n\left(n+1\right)}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}\)
\(=1-\frac{1}{\sqrt{121}}=\frac{10}{11}\)
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Bài 1: chắc lại phải "liên hợp" gì đó rồi:V
\(\sqrt{2009}-\sqrt{2008}=\frac{1}{\sqrt{2009}+\sqrt{2008}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Đó \(\sqrt{2009}+\sqrt{2008}>\sqrt{2007}+\sqrt{2006}\)
Nên \(\sqrt{2009}-\sqrt{2008}< \sqrt{2007}-\sqrt{2006}\)
Tổng quát ta có bài toán sau, với So sánh \(\sqrt{n}-\sqrt{n-1}\text{ và }\sqrt{n-2}-\sqrt{n-3}\)
Với \(n\ge3\). Lời giải xin mời các bạn:)
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Câu a)
Có: \(A=\sqrt{2009}-\sqrt{2008}\Leftrightarrow A^2=1-2\sqrt{2009\cdot2008}\)
\(B=\sqrt{2007}-\sqrt{2006}\Rightarrow B^2=1-2\sqrt{2007\cdot2006}\)
Đương nhiên: \(2\sqrt{2009\cdot2008}>2\sqrt{2006\cdot2007}\)
Suy ra: \(A< B\)
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Chứng minh rằng \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}< 1\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
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