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a) \(\dfrac{2007.2008-1}{2007.2008}\) và \(\dfrac{2008.2009-1}{2008.2009}\)
\(\dfrac{2007.2008-1}{2007.2008}=1+\dfrac{1}{2007.2008}\)
\(\dfrac{2008.2009-1}{2008.2009}=\dfrac{1}{2008.2009}\)
Vì \(\dfrac{1}{2007.2008}\)>\(\dfrac{1}{2008.2009}\) nên \(\dfrac{2007.2008-1}{2007.2008}\)>\(\dfrac{2008.2009-1}{2008.2009}\)
2007/2007.2008 và 2008/2008.2009
Ta có 2007/2007.2008=1/2008
2008/2008.2009=1/2009
Vì: 1/2008>1/2009
Nên: 2007/2007.2008>2008/2009
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
a) 2 phân số trên bằng nhau vì khi rút gọn \(\dfrac{6}{-27}\)với -3 ta được \(\dfrac{-2}{9}\)
=>\(\dfrac{-2}{9}=\dfrac{6}{-27}\)
b)-1/-5 và 4/25
-1/-5=-25/-125
4/25=-20/-125
=>-1/-5>4/25
a)
Ta có: \(BCNN\left( {10,15} \right) = 30\) nên
\(\begin{array}{l}\dfrac{7}{{10}} = \dfrac{{7.3}}{{10.3}} = \dfrac{{21}}{{30}}\\\dfrac{{11}}{{15}} = \dfrac{{11.2}}{{15.2}} = \dfrac{{22}}{{30}}\end{array}\)
Vì \(21 < 22\) nên \(\dfrac{{21}}{{30}} < \dfrac{{22}}{{30}}\) do đó \(\dfrac{7}{{10}} < \dfrac{{11}}{{15}}\).
b)
Ta có: \(BCNN\left( {8,24} \right) = 24\) nên
\(\dfrac{{ - 1}}{8} = \dfrac{{ - 1.3}}{{8.3}} = \dfrac{{ - 3}}{{24}}\)
Vì \( - 3 > - 5\) nên \(\dfrac{{ - 3}}{{24}} > \dfrac{{ - 5}}{{24}}\) do đó \(\dfrac{{ - 1}}{8} > \dfrac{{ - 5}}{{24}}\).
`M=(10^25+1)/(10^26+1)`
`=>10M=(10^26+10)/(10^26+1)=1+9/(10^26+1)``
`CMTT:10N=1+9/(10^27+1)`
Vì `1/(10^26+1)>1/(10^27+1)`
`=>9/(10^26+1)>9/(10^27+1)`
`=>1+9/(10^26+1)>1+9/(10^27+1)`
`=>10M>10N=>M>N`
a: \(\dfrac{-7}{6}=\dfrac{-7\cdot3}{6\cdot3}=\dfrac{-21}{18}\)
\(\dfrac{-11}{9}=\dfrac{-11\cdot2}{9\cdot2}=\dfrac{-22}{18}\)
mà -21>-22
nên \(-\dfrac{7}{6}>-\dfrac{11}{9}\)
b: \(\dfrac{5}{-7}=\dfrac{-5}{7}=\dfrac{-5\cdot5}{7\cdot5}=\dfrac{-25}{35}\)
\(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)
mà -25>-28
nên \(\dfrac{5}{-7}>\dfrac{-4}{5}\)
c: \(\dfrac{-8}{7}< -1\)
\(-1< -\dfrac{2}{5}\)
Do đó: \(-\dfrac{8}{7}< -\dfrac{2}{5}\)
d: \(-\dfrac{2}{5}< 0\)
\(0< \dfrac{1}{3}\)
Do đó: \(-\dfrac{2}{5}< \dfrac{1}{3}\)
a)
\(A=\dfrac{2007.2008-1}{2007.2008}=1-\dfrac{1}{2007.2008}\)
\(B=\dfrac{2008.2009-1}{2008.2009}=1-\dfrac{1}{2008.2009}\)
\(A-B=\dfrac{1}{2008.2009}-\dfrac{1}{2007.2008}< 0\Rightarrow A< B\)
b)
\(\left\{{}\begin{matrix}A=\dfrac{25}{43}\\B=\dfrac{10}{27}\end{matrix}\right.\) ; \(\dfrac{A}{B}=\dfrac{25}{43}.\dfrac{27}{10}=\dfrac{5.27}{43.2}>\dfrac{54.2}{43.2}>1\Rightarrow A>B\)
thanks b