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\(P=\dfrac{10^{2010}-1}{10^{2011}-1}\)
\(\Rightarrow10P=\dfrac{10\left(10^{2010}-1\right)}{10^{2011}-1}=\dfrac{10^{2011}-10}{10^{2011}-1}=\dfrac{10^{2011}-1-9}{10^{2011}-1}=1-\dfrac{9}{10^{2011}-1}\)
\(Q=\dfrac{10^{2009}-1}{10^{2010}-1}\)
\(\Rightarrow10Q=\dfrac{10\left(10^{2009}-1\right)}{10^{2010}-1}=\dfrac{10^{2010}-10}{10^{2010}-1}=\dfrac{10^{2010}-1-9}{10^{2010}-1}=1-\dfrac{9}{10^{2010}-1}\)Mà\(\dfrac{9}{10^{2011}-1}< \dfrac{9}{10^{2010}-1}\)
\(\Rightarrow1-\dfrac{9}{10^{2011}-1}>1-\dfrac{9}{10^{2010}-1}\)
Hay \(10P>10Q\Rightarrow P>Q\)
Do 20092010- 2 < 20092011- 2 ⇒ B < 1
\(B=\frac{2009^{2010}-2}{2009^{2011}-2}<\frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(1+2009^{2009}\right)}{2009\left(1+2009^{2010}\right)}\)
\(=\frac{2009^{2009}+1}{2009^{2010}+1}=A\Rightarrow\)B < A
Ta có: \(A=\frac{10^{18}+1}{10^{19}+1}>\frac{10.\left(10^{17}+1\right)}{10.\left(10^{18}+1\right)}=\frac{10^{17}+1}{10^{18}+1}\)
Vậy A < B
Vì \(\frac{10^{18}+1}{10^{19}+1}< 1\Rightarrow B=\frac{10^{18}+1}{10^{19}+1}< \frac{10^{18}+1+9}{10^{19}+1+9}\)
\(\Rightarrow B< \frac{10^{18}+10}{10^{19}+10}\)
\(\Rightarrow B< \frac{10\left(10^{17}+1\right)}{10\left(10^{18}+1\right)}\)
\(\Rightarrow B< \frac{10^{17}+1}{10^{18}+1}\)
\(\Rightarrow B< A\)
Vậy A > B.
So sánh:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}\) và \(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}\)
Ta có:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}=-\frac{1}{10^{2011}}\left(\frac{9}{10}+19\right)=-\frac{1}{10^{2011}}.\frac{199}{10}\)
\(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}=-\frac{1}{10^{2011}}\left(9+\frac{19}{10}\right)=-\frac{1}{10^{2011}}.\frac{109}{10}\)
Vì \(\frac{199}{10}>\frac{109}{10}\Rightarrow\frac{1}{10^{2011}}.\frac{199}{10}>\frac{1}{10^{2011}}.\frac{109}{10}\Rightarrow-\frac{1}{10^{2011}}.\frac{199}{10}< -\frac{1}{10^{2011}}.\frac{109}{10}\)
Vậy nên A<B
1/ (69.210+1210)+(219.273+15.49.94) = 29.39.210+310.220+219.39+5.3.218.38 = 219.39+310.220+219.39+5.218.39
= 218.39(2+3.22+5)=19.218.39
ta có quy đồng B ta dc(-9x10^2018-19x10^2019)/(10^2019x10^2018)
tương tự với C ta có (-19x10^2018-9x10^2019)/(10^2019x10^2018)
sau khi quy đồng ta thấy mẫu của B và C giống nhau từ đó ta so sánh tử số của B và C
tử số của B=10^2018x(-9-19x10)=10^2018x-199
C=10^2018x(-19-9x10)=10^2018x-109
ta thấy -199<-109=>B<C (dpcm)
Ta có: \(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-9}{10^{2010}}-\frac{9}{10^{2011}}-\frac{10}{10^{2011}}\)
\(=\frac{-9}{10^{2010}}-\frac{9}{10^{1011}}-\frac{1}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}\)
Ta thấy : \(\frac{10}{10^{2010}}< \frac{19}{10^{2010}}\Rightarrow\frac{-10}{10^{2010}}>\frac{-19}{10^{2010}}\)
\(\Rightarrow\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}>\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
Hay \(A>B\)
Vậy ...