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Ta có:
A = 2017 2018 + 2018 2019 > 2010 2011 + 2011 2012 = 2010 + 2011 2012 > 2010 + 2011 2011 + 2012 = B
Vậy A > B
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b, \(\dfrac{56\cdot2+56\cdot3}{15\cdot56}=\dfrac{56\cdot\left(2+3\right)}{15\cdot56}\)
=\(\dfrac{56\cdot5}{15\cdot56}=\dfrac{1}{3}\)
c, \(\dfrac{20082008}{20112011}=\dfrac{2008\cdot10001}{2011\cdot10001}=\dfrac{2008}{2011}\)
d, \(\dfrac{2+2}{217\cdot32+217+33}=\dfrac{4}{217\cdot\left(32+33\right)}\)
=\(\dfrac{4}{217\cdot65}=\dfrac{4}{14105}\)
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\(\frac{199.......9}{999.......5}=\frac{199.......9:199.......9}{999.......5:199........9}=\frac{1}{5}\)
\(\frac{121212}{424242}=\frac{121212:10101}{424242:10101}=\frac{12}{42}\)
\(\frac{187187187}{221221221}=\frac{187187187:1001001}{221221221:1001001}=\frac{187}{221}\)
\(\frac{20082008}{20112011}=\frac{20082008:10001}{20112011:10001}=\frac{2008}{2011}\)
\(\frac{199...9}{99...95}\)=\(\frac{199...9:199...9}{99...95:199...9}\)\(\frac{\left(20so9\right)}{\left(20so9\right)}\)
=\(\frac{1}{5}\)
A = 20112011 + 2/20112011 - 1
A=(20112011-1+3)/(20112011 - 1)=1 + 3/(20112011 - 1)
B=20112011/20112011 - 3.
B=(20112011-3+3)/(20112011 - 3)=1 + 3/(20112011 - 3)
Vì 3/(20112011 - 1) < 3/(20112011 - 3) {Hai phân số cùng tử,phân số nào mẫu lớn hơn thì nhỏ hơn}
Nên A<B
Ta có: \(2011>2011^{2011}-3\)
\(\Rightarrow\frac{2011^{2011}}{2011^{2011}-3}>1\)
\(\Rightarrow B>\frac{2011^{2011}+2}{2011^{2011}-3+2}=\frac{2011^{2011}+2}{2011^{2011}-1}=A\)
Vậy \(B>A\)