Cho \(A=\frac{1}{1.18}+\frac{1}{2.19}+\frac{1}{3.20}+...+\frac{1}{2000.2017}\)
\(B=\frac{1}{1.2001}+\frac{1}{2.2002}+\frac{1}{3.2003}+...+\frac{1}{17.2017}\)
Tính A/B
Cho \(A=\dfrac{1}{1.18}+\dfrac{1}{2.19}+\dfrac{1}{3.20}+...+\dfrac{1}{2000.2017}\)
\(B=\dfrac{1}{1.2001}+\dfrac{1}{2.2002}+\dfrac{1}{3.2003}+...+\dfrac{1}{17.2017}\)
Tính A/B
Cho \(A=133.\left(\frac{1}{1.1996}+\frac{1}{1.1997}+...+\frac{1}{17.2012}\right)\)
\(B=\frac{17}{15}\left(\frac{1}{1.18}+\frac{1}{2.19}+...+\frac{1}{1995.2012}\right)\)
So sánh A, B
Giúp mk vs
Cảm ơn nhiều ạ
Ta có A = \(133\left(\frac{1}{1.1996}+\frac{1}{2.1997}+...+\frac{1}{17.2002}\right)\)
=> 1995A = \(133\left(\frac{1995}{1.1996}+\frac{1995}{2.1997}+...+\frac{1995}{17.2002}\right)\)
=> 1995A = \(133\left(1-\frac{1}{1996}+\frac{1}{2}-\frac{1}{1997}+...+\frac{1}{17}-\frac{1}{2002}\right)\)
=> 1995A = \(133\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2002}\right)\right]\)
=> A = \(\frac{1}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2002}\right)\right]\)(1)
Lại có B = \(\frac{17}{15}\left(\frac{1}{1.18}+\frac{1}{2.19}+...+\frac{1}{1995.2012}\right)\)
=> 17B = \(\frac{17}{15}\left(\frac{17}{1.18}+\frac{17}{2.19}+...+\frac{17}{1995.2012}\right)\)
=> 17B = \(\frac{17}{15}\left(1-\frac{1}{18}+\frac{1}{2}-\frac{1}{19}+...+\frac{1}{1995}-\frac{1}{2012}\right)\)
=> 17B = \(\frac{17}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{1995}\right)-\left(\frac{1}{18}+\frac{1}{19}+...+\frac{1}{2012}\right)\right]\)
=> 17B = \(\frac{17}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2012}\right)\right]\)
=> B = \(\frac{1}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2012}\right)\right]\)(2)
Từ (1) và (2) => A = B
Cho
\(A=124.\left(\frac{1}{1.1985}+\frac{1}{2.1986}+...+\frac{1}{16.2000}\right)\)
\(B=\frac{1}{1.17}+\frac{1}{1.18}+\frac{1}{3.19}+...+\frac{1}{1984.2000}\)
So sánh A và B
a, Cho a + b + c =0 chứng minh:
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\)
b, Tính
\(A=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{399^2}+\frac{1}{400^2}}\)
Mình giúp phần a thôi, phần b chir là áp dụng không có gì khó cả.
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\left(a+b+c=0\right)\)
\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\left(đpcm\right)\)
b, \(A=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{399^2}+\frac{1}{400^2}}\)
\(A=\sqrt{\frac{1}{1^2}+\frac{1}{1^2}+\frac{1}{\left(-2\right)^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{\left(-3\right)^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{399^2}+\frac{1}{\left(-400\right)^2}}\)
có 1 + 1 - 2 = 1 + 2 - 3 = ... + 1 + 399 - 400 = 0
nên theo câu a ta có :
\(A=\left|1+\frac{1}{1}-\frac{1}{2}\right|+\left|1+\frac{1}{2}-\frac{1}{3}\right|+...+\left|1+\frac{1}{399}-\frac{1}{400}\right|\)
A = 1 + 1 -1/2 + 1 + 1/2 - 1/3 + 1 + 1/3 - 1/4 + ... + 1 + 1/399 - 1/400
= 400 1/400
= 159999/400
Bạn ơi cho mình hỏi áp dụng như lào vậy???
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{9.10}\) và \(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\). Tính A : B
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{9.10}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\)
\(\Rightarrow A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
\(\Rightarrow A=\left(\frac{1}{6}+\frac{1}{10}\right)+\left(\frac{1}{7}+\frac{1}{9}\right)+\frac{1}{8}\)
\(\Rightarrow A=\left(\frac{10}{6.10}+\frac{6}{6.10}\right)+\left(\frac{9}{7.9}+\frac{7}{7.9}\right)+\frac{8}{8.8}\)
\(\Rightarrow A=\frac{16}{6.10}+\frac{16}{7.9}+\frac{8}{8.8}\)
\(\Rightarrow A=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)\)
Ta lại có:
\(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\)
\(\Rightarrow B=\left(\frac{1}{6.10}+\frac{1}{6.10}\right)+\left(\frac{1}{7.9}+\frac{1}{7.9}\right)+\frac{1}{8.8}\)
\(\Rightarrow B=\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\)
Vậy :
\(A:B=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right):\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)=8\)
Vậy \(A:B=8\)
Cho ab=1 và a+b≠0. Tính
\(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Cho a,b thỏa ab=1; a+b\(\ne\) 0 Tính
\(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{1}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)
cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{107}+\frac{1}{108}\);\(B=\frac{108}{1}+\frac{107}{2}+\frac{106}{3}+...+\frac{2}{107}+\frac{1}{108}\).tính\(\frac{A}{B}\)
a)A=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
b)A =\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\)và B = \(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{3}{2016}\)
Tính \(\frac{B}{A}\)
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)