Tính \(\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}...\frac{2011^2}{2012^2-1}\)
Bài 3 : a) Tính
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) Tính :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\frac{1}{2011}}\)
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
1.Tìm tất cả các số tự nhiên n thỏa mãn:
\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right)^{2n -1}+n.2^n=8192\)
2. So sánh A và B biết:
\(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{5.6}+...+\frac{2011}{1999.2000}\)
\(B=\frac{2012}{1001}+\frac{2012}{1002}+\frac{2012}{1003}+...+\frac{2012}{2000}\)
3. Tính \(\left(S-P\right)^{2016}\) biết:\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
4.Tìm x:
a) \(-1\frac{1}{56}:\left(\frac{1}{8}-\frac{1}{7}\right)-\frac{22}{\left|2.x-0,5\right|}=-1\frac{1}{30}:\left(\frac{1}{5}-\frac{1}{6}\right)\)
b) \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}....\frac{30}{62}.\frac{31}{64}=2^x\)
c) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)
Tính giá trị biểu thức :
\(A=\frac{\frac{1}{2013}+\frac{2}{2012}+\frac{3}{2011}+...+\frac{2011}{3}+\frac{2012}{2}+\frac{2013}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
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1/ Chứng tỏ rằng \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}<1\)
2/ Chứng tỏ rằng \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}<1\)
3/ Rút gọn biểu thức \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
4/ Tính nhanh\(\frac{\frac{4}{2010}+\frac{4}{2011}-\frac{4}{2012}}{\frac{5}{2010}+\frac{5}{2011}-\frac{5}{2012}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
GIÚP ĐƯỢC CÂU NÀO THÌ GIÚP NHÉ, MÌNH TICK CHO
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
a)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)=\(\frac{99}{100}<1\)
Tính: \(\frac{1^2}{2^2-1}\). \(\frac{3^2}{4^2-1}\).\(\frac{5^2}{6^2-1}\) ..........................\(\frac{2011^2}{2012^2-1}\)
\(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+.....+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}\)
rút gọn giúp mình với
Xét biểu thức phụ : \(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
tính
\(B=\frac{6}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:1\frac{1}{2011}:\left(-1\frac{1}{2012}\right)\)
Tính nhanh
\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+\frac{2010}{5}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
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Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
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đùa thôi đáp án: 2015 nha bn
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