Chào bạn, bạn hãy theo dõi câu trả lời của mình nhé! 

a) Ta có : 

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{151}=3^{150}\cdot3=\left(3^2\right)^{75}\cdot3=9^{75}\cdot3\)

Mà \(9^{75}>8^{75}=>9^{75}\cdot3>8^{75}=>3^{151}>2^{225}\)

b) Nhân cả vế A lẫn vế B với 10²⁰⁰⁵, ta có : 

\(10^{2005}A=-7+\frac{-15}{10}=\frac{-70}{10}+\frac{-15}{10}=\frac{-85}{10}\)

\(10^{2005}B=-15+\frac{-7}{10}=\frac{-150}{10}+\frac{-7}{10}=\frac{-157}{10}\)

Mà \(\frac{-85}{10}>\frac{-157}{10}=>10^{2005}A>10^{2005}B\)

\(=>A>B\)

Chúc bạn học tốt!

C1: Ta có 8/9<1 mà 9/8>1 nên 8/9,9/8

C2: Quy đồng mẫu số 2 Phân số 8/9 và 9/8 ta được:8.8/9.8=64/72

                                                                                          9.9/8.9=81/72

Vì 81/72>64/72 nên 9/8>8/9

Cách 1:

Ta thấy \(\frac{8}{9}<1\) và \(\frac{9}{8}>1\)

=>\(\frac{8}{9}<\frac{9}{8}\)

Cách 2: 

\(\frac{8}{9}=\frac{8X8}{9X8}=\frac{64}{72}\)

\(\frac{9}{8}=\frac{9X9}{8X9}=\frac{81}{72}\)

Mà \(\frac{64}{72}<\frac{81}{72}\)

Vậy \(\frac{8}{9}<\frac{9}{8}\)

Ta có:

Q=2010/2011+2012+2013+2011/2011+2012+2013+2012/2011+2012+2013

Mà 2010/2011+2012+2013<2010/2011

      2011/2011+2012+2013<2011/2012

      2012/2011+2012+2013<2012/2013

=>Q<P

Có : \(\frac{2011}{2012}=\frac{2012-1}{2012}=1-\frac{1}{2012}\)

Có : \(\frac{2012}{2013}=\frac{2013-1}{2013}=1-\frac{1}{2013}\)

Có : \(\frac{2013}{2011}=\frac{2011+2}{2011}=1+\frac{2}{2011}\)

Cộng vế với vế ta có : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{2}{2011}=1+1+1-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)=3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)\)

Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}>0\) nên \(3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)<3\)

Vậy \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}<3\)

ta co:B=20¹⁰-1/20¹⁰-3>1

=>B>20¹⁰-1+2/20¹⁰-3+2=20¹⁰+1/20¹⁰-1=A

vay A<B

dễ thấy B=\(\frac{2015+2016}{2016+2017}\)<1

A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)=1-\(\frac{1}{2016}\)+1-\(\frac{1}{2017}\)=(1+1)-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))=2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))

vì (\(\frac{1}{2016}\)+\(\frac{1}{2017}\))<0,5+0,5=1 suy ra 2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))>1 mà b<1suy ra A>B

Ta thấy: B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)

              A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)

Mà\(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)

Suy ra: \(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)>\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)=\(\frac{2015+2016}{2016+2017}\)

               Hay A>B

Ta có: 
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80 
và 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80) 
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60 
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60 
và 1/61> 1/62> ... >1/79> 1/80 
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80 
Vậy 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12 

Giải:

\(\frac{13}{20}=\frac{13.101}{20.101}=\frac{1313}{2020}\)

\(\frac{100}{101}=\frac{100.20}{101.20}=\frac{2000}{2020}\)

Vì \(1313<2000\Rightarrow\frac{1313}{2020}=\frac{2000}{2020}\Rightarrow\frac{13}{20}<\frac{100}{101}\)

Chúc bạn học tốt!

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