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Bài 1
\(\frac{2017}{2018}+\frac{2018}{2019}\)và \(\left(\frac{2017+2018}{2018+2019}\right)\)mk chữa lại đề luôn đó
Ta tách :
\(\frac{2017}{\left(2018+2019\right)+2018}\)
đến đây ta tách
\(\frac{2017}{2018+2019}< \frac{2017}{2018}\)
vậy....
mấy câu khác tương tự
2) \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{2}{2003}+\frac{2}{2004}+\frac{2}{2005}}\)
= \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{2.\frac{1}{2003}+2.\frac{1}{2004}+2.\frac{1}{2005}}\)
=\(\frac{1\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}{2.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}\)
= \(\frac{1}{2}\)
3) \(2013+\left(\frac{2013}{1+2}\right)+\left(\frac{2013}{1+2+3}\right)+...+\left(\frac{2013}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2025078}\right)\)
= \(2013.2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4050156}\right)\)
=\(4026.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
= \(4026.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
= \(4026.\left(1-\frac{1}{2013}\right)\)
= \(4026.\frac{2012}{2013}\)
=\(4024\)
a)2012/2013<2013/2014
b)2003x........< 2007x......
cho mình nha
Có : 2004A = 2004^2004+2004/2004^2004+1 = 1 + 2003/2004^2004+1
2004B = 2004^2005+2004/2004^2005+1 = 1 + 2003/2004^2005+1 < 1 + 2003/2004^2004+1 = 2014A
=> A > B
Tk mk nha
Câu hỏi của linh phạm - Toán lớp 6 - Học toán với OnlineMath
\(A=\frac{2003\cdot2004-1}{2003\cdot2004}=1-\frac{1}{2003\cdot2004}\)
\(B=\frac{2004\cdot2005-1}{2004\cdot2005}=1-\frac{1}{2004\cdot2005}\)
Vì 1 = 1 và \(\frac{1}{2003\cdot2004}>\frac{1}{2004\cdot2005}\) nên A > B
Vậy A > B
Chắc sai =))
\(A=\frac{2003\cdot2004-1}{2003\cdot2004}=\frac{2003\cdot2004}{2003\cdot2004}-\frac{1}{2003\cdot2004}=1-\frac{1}{2003\cdot2004}\)
\(B=\frac{2004\cdot2005-1}{2004\cdot2005}=\frac{2004\cdot2005}{2004\cdot2005}-\frac{1}{2004\cdot2005}=1-\frac{1}{2004\cdot2005}\)
có : \(\frac{1}{2003\cdot2004}>\frac{1}{2004\cdot2005}\)
\(\Rightarrow1-\frac{1}{2003\cdot2004}< 1-\frac{1}{2004\cdot2005}\)
\(\Rightarrow A< B\)
a, Ta có: \(\frac{2012.2013}{2012.2013+1}< 1< \frac{2013}{2012}\)
\(\Rightarrow\frac{2012.2013}{2012.2013+1}< \frac{2013}{2012}\)
b, \(A=\frac{2003.2004-1}{2003.2004}=1-\frac{1}{2003.2004}\)
\(B=\frac{2004.2005-1}{2004.2005}=1-\frac{1}{2004.2005}\)
Ta có: \(2003.2004< 2004.2005\)
\(\Rightarrow\frac{1}{2003.2004}>\frac{1}{2004.2005}\)
\(\Rightarrow1-\frac{1}{2003.2004}< 1-\frac{1}{2004.2005}\)
\(\Rightarrow A< B\)