so sánh \(\frac{-1000}{999}\)và \(\frac{-2005}{2006}\)
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a) \(\frac{999}{10000}=\frac{99,9}{1000}>\frac{99}{100}\)
=> kết luận
b) \(1-\frac{97}{99}=\frac{2}{99}>1-\frac{98}{100}=\frac{2}{100}\)
\(\Rightarrow\frac{97}{99}< \frac{98}{100}\)
=> kết luận
12/11 lớn hơn vì 12/11 lớn hơn 1 còn 2005/2006 bé hơn 1
Ta có A=1/102005(-7-15/10)=1/102005.(-8,5) (1) B=1/102005(-15-7/10)=1/102005.(-15,7) (2) (1)(2)-> A>B
C=1/2.(3/60.63+....+3/117.120)+1/1003
C=1/2.(1/60-1/63+....+1/117-1/120)+1/1003
....còn lại tự làm nha, bài còn lại cũng tương tự
Ta có:
\(C=\dfrac{2}{60.63}+\dfrac{2}{63.66}+...+\dfrac{2}{117.120}+\dfrac{2}{2006}\)
\(C=2\left(\dfrac{1}{60.63}+\dfrac{1}{63.66}+...+\dfrac{1}{117.120}\right)+\dfrac{2}{2006}\)
\(C=2.\dfrac{1}{3}\left(\dfrac{3}{60.63}+\dfrac{3}{63.66}+...+\dfrac{3}{117.120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}.\dfrac{1}{120}+\dfrac{2}{2006}\)
\(C=\dfrac{1}{180}+\dfrac{2}{2006}\)
Ta lại có:
\(D=\dfrac{5}{40.44}+\dfrac{5}{44.48}+...+\dfrac{5}{76.80}+\dfrac{5}{2006}\)
\(D=5\left(\dfrac{1}{40.44}+\dfrac{1}{44.48}+...+\dfrac{1}{76.80}\right)+\dfrac{5}{2006}\)
\(D=5.\dfrac{1}{4}\left(\dfrac{4}{40.44}+\dfrac{4}{44.48}+...+\dfrac{4}{76.80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}.\dfrac{1}{80}+\dfrac{5}{2006}\)
\(D=\dfrac{1}{64}+\dfrac{5}{2006}\)
Vì \(\dfrac{1}{180}< \dfrac{1}{64}\)
\(\dfrac{2}{2006}< \dfrac{5}{2006}\)
\(\Rightarrow\dfrac{1}{180}+\dfrac{2}{2006}< \dfrac{1}{64}+\dfrac{5}{2006}\)
\(\Rightarrow C< D\)
dở ẹt nhu cu net ma ko biet lamb tao hoc lop mau giao tao cung biet tra loi dung la ngu
Ta có: 1-2004/2005=1/2005
1-2005/2006=1/2006
Mà 1/2005>1/2006 (sử dụng phần bù)
=>2004/2005<2005/2006
Vì 2003 / 2004 < 1 và 1 < 2006 / 2005 => 2003 / 2004 < 2006 / 2005
Trả lời:
Vì \(\frac{2003}{2004}< 1;1< \frac{2006}{2005}\)
\(\Rightarrow\frac{2003}{2004}< \frac{2006}{2005}\)
Vậy \(\frac{2003}{2004}< \frac{2006}{2005}\)
\(A=\frac{100^{2007}+1}{100^{2008}+1}\Rightarrow100.A=\frac{100^{2008}+100}{100^{2008}+1}=\frac{100^{2008}+1+99}{100^{2008}+1}=1+\frac{99}{100^{2008}+1}\)
\(B=\frac{100^{2006}+1}{100^{2007}+1}\Rightarrow100.B=\frac{100^{2007}+100}{100^{2007}+1}=\frac{100^{2007}+1+99}{100^{2007}+1}=1+\frac{99}{100^{2007}+1}\)
Vì \(\frac{99}{100^{2007}+1}>\frac{99}{100^{2008}+1};1=1\Rightarrow1+\frac{99}{100^{2007}+1}>1+\frac{99}{100^{2008}+1}\)hay \(A>B\)
Vậy \(A>B\)
Ta có:\(\frac{-1000}{999}\)<-1
\(\frac{-2005}{2006}\)<0
Vì -1<0 nên \(\frac{-1000}{999}\)<\(\frac{-2005}{2006}\)
Chúc bạn học tốt!