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!

bạn có thể vào để hỏi nhé !

Chúc bạn học tốt !

A> \(\frac{10^n-2-2}{10^n-1-2}=\frac{10^n-4}{10^n-3}=B\)

=> A>B

A=\(\frac{2014}{2014^a}+\frac{2014}{2014^b}\)=B=\(\frac{2013}{2015^a}\)+\(\frac{2015}{2013^b}\)

Ta có: 2014/\(2014^a\)+2014/2014^b= 2013/2014^a + 1/2014^a +2015/2014^a - 1/2014^a

                                                        =(2013/2014^a + 2015/2014^b) + ( 1/2014^a + 1/2014^b)

                                                       =                   B                                 + (1/2014^a + 1/2014^b)

   *Nếu a=b thì A=B

   *Nếu a>b thì (1/2014^a + 1/2014^b) >0

                      \(\Rightarrow\) A< B

   *Nếu a<b thì (1/2014^a + 1/2014^b)>0

                     \(\Rightarrow\) 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 

Ta có :

\(\frac{1+2+3+...+a}{a}<\frac{1+2+3+...+b}{b}\)

\(\Leftrightarrow\frac{a\left(a+1\right)}{a}<\frac{b\left(b+1\right)}{b}\)

<=> a + 1 < b + 1

<=> a < b

có 1+2+3+...+a/a<1+2+3+...+b/b

=>(a+1)(a-1+1):2/a<(b+1)(b-1+1):2/b

<=>(a+1)a:2/a<(b+1)b;2/b

<=>a+1<b+1

<=>a<b

vậy a<b

B=\(\frac{2011^{10}-1}{2011^{10}-3}\) <1 => \(\frac{2011^{10}-1}{2011^{10}-3}\) < \(\frac{2011^{10}-1+2}{2011^{10}-3+2}\) = \(\frac{2011^{10}+1}{2011^{10}-1}\) = A

=> B<A

Cảm ơn bạn nhiều nha giải ra lại thấy dễ ak

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

Bài 3:

\(\left(\dfrac{1}{32}\right)^7=\dfrac{1^7}{32^7}=\dfrac{1}{32^7}=\dfrac{1}{\left(2^5\right)^7}=\dfrac{1}{2^{35}}\\ \left(\dfrac{1}{16}\right)^9=\dfrac{1^9}{16^9}=\dfrac{1}{16^9}=\dfrac{1}{\left(2^4\right)^9}=\dfrac{1}{2^{36}}\)

Vì \(2^{35}< 2^{36}\) nên \(\dfrac{1}{2^{35}}>\dfrac{1}{2^{36}}\) hay \(\left(\dfrac{1}{32}\right)^7>\left(\dfrac{1}{16}\right)^9\)