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Câu hỏi của Nguyễn Tuấn Minh - Toán lớp 6 - Học toán với OnlineMath
Vì \(2014.2015=2014.2015\)nên \(2014.2015-1< 2014.2015\)1 đơn vi
Vì \(2015.2016=2015.2016\)nên \(2015.2016-1< 2015.2016\)1 đơn vị
Ta có :
\(1-M=1-\frac{2014.2015-1}{2014.2015}=\frac{1}{2014.2015}\)
\(1-N=1-\frac{2015.2016-1}{2015.2016}=\frac{1}{2015.2016}\)
Vì \(2015=2015\)nên \(2014.2015< 2015.2016\)
Vì \(\frac{1}{2014.2015}>\frac{1}{2015.2016}\)( do \(2014.2015< 2015.2016\))
Nên \(N>M\)
Vậy \(N>M\)
a) So sánh \(\frac{461}{456}\) và \(\frac{128}{123}\):
\(\frac{461}{456}\) = \(\frac{456+5}{456}=1+\frac{5}{456};\frac{128}{123}=\frac{123+5}{123}=1+\frac{5}{123}\)
Vì \(\frac{1}{456}<\frac{1}{123}\Rightarrow\frac{5}{456}<\frac{5}{123}\Rightarrow\frac{461}{456}<\frac{128}{123}\Rightarrow\frac{456}{461}>\frac{123}{128}\)(Ta có tính chất: nếu 0< a< b thì 1/a > 1/b)
b) \(\frac{2014.2015-1}{2014.2015}=1-\frac{1}{2014.2015}\) ; \(\frac{2015.2016-1}{2015.2016}=1-\frac{1}{2015.2016}\)
vì 2014.2015 < 2015.2016 nên \(\frac{1}{2014.2015}>\frac{1}{2015.2016}\Rightarrow1-\frac{1}{2014.2015}<1-\frac{1}{2015.2016}\Rightarrow\frac{2014.2015-1}{2014.2015}<\frac{2015.2016-1}{2015.2016}\)
\(P=\dfrac{2013.2014-1007.4030}{2014^2-2011.2014}\)
\(P=\dfrac{2013.2014-1007.2.2015}{2014^2-2011.2014}\)
\(P=\dfrac{2013.2014-2014.2015}{2014^2-2011.2014}\)
\(P=\dfrac{2014.\left(2013-2015\right)}{2014.\left(2014-2011\right)}\)
\(P=-\dfrac{2}{3}\)
\(Q=-\dfrac{214263}{142862}=-\dfrac{214263:71421}{142862:71421}=-\dfrac{3}{2}\)
Vì \(-\dfrac{2}{3}>-\dfrac{3}{2}\)nên P>Q
\(P=\dfrac{2013.2014-1007.4030}{2014^2-2011.2014}\)
\(P=\dfrac{2013.2014-1007.4030}{2014.2014-2011.2014}\)
\(P=\dfrac{2013.2.1007-1007.4030}{2014\left(2014-2011\right)}\)
\(P=\dfrac{4026.1007-1007.4030}{2014.3}\)
\(=\dfrac{1007.\left(4026-4030\right)}{1007.2.3}\)
\(=\dfrac{1007.-4}{1007.6}=\dfrac{-4}{6}=\dfrac{-2}{3}\)
\(Q=\dfrac{214263}{142842}=-\dfrac{3}{2}\)
\(-\dfrac{3}{2}< -\dfrac{2}{3}\Rightarrow P>Q\)
\(A=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995\left(1993+1\right)-1}{1993.1995+1994}=\dfrac{1995.1993+1995-1}{1993.1995+1994}=\dfrac{1995.1993+1994}{1995.1993-1994}=1\)\(B=\dfrac{2004.2004+3006}{2005.2005-1003}=\dfrac{2004.2004+2004.1+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2004.2005+2005-1003}=\dfrac{2004.2005+1002}{2004.2005+1002}=1\)\(C=\dfrac{2010.2011-1}{2009.2011+2010}=\dfrac{2009.2011+2011-1}{2009.2011+2010}=\dfrac{2019.2011+2010}{2009.20011+2010}=1\)\(D=\dfrac{2014.2015-1}{2013.2015+2013}=\dfrac{2013.2015+2014-1}{2013.2015+2013}=\dfrac{2013.2015+2013}{2013.2015+2013}=1\)
Câu 1 nhầm đề nha bạn mình sửa:
\(\dfrac{1995.1994-1}{1993.1995+1994}\)
\(=\dfrac{1995.\left(1993+1\right)-1}{1993.1995+1994}\)
\(=\dfrac{1995.1993+1995-1}{1993.1995+1994}\)
\(=\dfrac{1993.1995+1994}{1993.1995+1994}\)
\(=1\)
Câu 2: \(\dfrac{2004.2004+3006}{2005.2005-1003}\)
\(=\dfrac{2004.2004+2004+1002}{\left(2004+1\right).\left(2004+1\right)-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1002}\)
\(=1\)
Câu 3:\(\dfrac{2010.2011-1}{2009.2011+2010}\)
\(=\dfrac{\left(2009+1\right).2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2010}{2009.2011+2010}\)
= 1
Câu 4:Nhầm để, sửa:
\(\dfrac{2014.2015-1}{2013.2015+2014}\)
\(=\dfrac{\left(2013+1\right).2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2014}{2013.2015+2014}\)
\(=1\)
1. Ta có: \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(m\in Z\right)\)
\(B=\dfrac{2016^{2016}}{2016^{2016}-3}>\dfrac{2016^{2016}+2}{2016^{2016}-3+2}=\dfrac{2016^{2016}+2}{2016^{2016}-1}=A\)
Vậy A > B
2. \(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+...+\dfrac{1}{1.2}\)
= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
= \(1-\dfrac{1}{2016}\)
=\(\dfrac{2015}{2016}\)
Ta có : \(A\text{=}\dfrac{2013.2014-1}{2013.2014}\text{=}\dfrac{2013.2014}{2013.2014}-\dfrac{1}{2013.2014}\text{=}1-\dfrac{1}{2013.2014}\)
\(B\text{=}\dfrac{2014.2015-1}{2014.2015}\text{=}\dfrac{2014.2015}{2014.2015}-\dfrac{1}{2014.2015}\text{=}1-\dfrac{1}{2014.2015}\)
\(Ta\) có : \(\dfrac{1}{2013.2014}>\dfrac{1}{2014.2015}\)
\(\Rightarrow A< B\)