E=\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{2005^2}+\frac{1}{2006^2}}\)
Rút gọn E
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Rút gọn E
E = \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2005^2}+\frac{1}{2006^2}}\)
Rút gọn
A=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}+2004\sqrt{2005}}\)
Ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Từ đó ta có
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)
\(=1-\frac{1}{\sqrt{2005}}=\frac{\sqrt{2005}-1}{\sqrt{2005}}\)
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CM \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2006\sqrt{2005}}< 2\)
bài 1: Cho a+b+c0Chứng minh đẳng thức sqrt{frac{1}{a^2}+frac{1}{b^2}+frac{1}{c^2}}left|frac{1}{a}+frac{1}{b}+frac{1}{c}right|Bài 2: Cho Afrac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+....+frac{1}{sqrt{2005}+sqrt{2006}}Bfrac{1}{sqrt{1}}+frac{1}{sqrt{2}}+frac{1}{sqrt{3}}+....+frac{1}{sqrt{2005}}a, Rút gọ Ab, Chứng minh B2left(sqrt{2006}-1right) Giúp mk vs ạ !!! Cô cho bài về để ôn thi học kì mà ko pic lm nà :((( !!!
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bài 1: Cho \(a+b+c=0\)Chứng minh đẳng thức
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
Bài 2: Cho \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{2005}+\sqrt{2006}}\)
\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2005}}\)
a, Rút gọ A
b, Chứng minh \(B>2\left(\sqrt{2006}-1\right)\)
Giúp mk vs ạ !!! Cô cho bài về để ôn thi học kì mà ko pic lm nà :((( !!!
1/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)
\(\Leftrightarrow\frac{a+b+c}{abc}=0\)(đúng)
Vậy ta có ĐPCM
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2/ \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2005}+\sqrt{2006}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2006}-\sqrt{2005}\)
\(=\sqrt{2006}-1\)
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b/ Ta có
\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}\)
\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\)
Áp dụng vài bài toán ta có
\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2005}}\)
\(>2.\sqrt{2}-2.\sqrt{1}+2.\sqrt{3}-2.\sqrt{2}+...+2.\sqrt{2006}-2.\sqrt{2005}\)
\(=2.\sqrt{2006}-2=2\left(\sqrt{2006}-1\right)\)
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Xem thêm câu trả lời
Rút gọn các biểu thức sau:
a,\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}\)
b,\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{2006}+\sqrt{2007}}\)
c,\(C=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
chứng tỏ rằng :
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3.\left(2-1\right)}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5\left(3-2\right)}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011\left(2006-2005\right)}\)
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011}\)
Nhận xét: (a-b)2 \(\ge\) 0 => a2 + b2 \(\ge\) 2ab
Áp dụng ta có: \(3=\left(\sqrt{2}\right)^2+\left(\sqrt{1}\right)^2\ge2.\sqrt{2}.\sqrt{1}\)
\(5=\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2\ge2.\sqrt{3}.\sqrt{2}\)
...
\(4011=\left(\sqrt{2006}\right)^2+\left(\sqrt{2005}\right)^2\ge2.\sqrt{2006}.\sqrt{2005}\)
=> \(VT
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Tính tổng \(S=\sqrt{1+\frac{1}{1^2}+\frac{1}{^{2^2}}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{^{3^2}}}+...+\sqrt{1+\frac{1}{2005^2}+\frac{1}{^{2006^2}}}\)
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2\left(1.\frac{1}{n}-1.\frac{1}{n+1}-\frac{1}{n}.\frac{1}{n+1}\right)=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\); vì \(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n\left(n+1\right)}=0\)
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{2005}-\frac{1}{2006}\right)\)
\(=2005+1-\frac{1}{2006}=2005\frac{2005}{2006}\)
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