\(10\equiv1\left(mod3\right)\Leftrightarrow10^{2013}\equiv1\left(mod3\right)\\ 2014\equiv1\left(mod3\right)\\ \Leftrightarrow10^{2013}-2014\equiv1-1=0\left(mod3\right)\\ \Leftrightarrow10^{2013}-2014⋮3\)

Mod là Gì vậy

\(\dfrac{1}{10}A=\dfrac{10^{2012}+1}{10^{2012}+10}=1-\dfrac{9}{10^{2012}+10}\)

\(\dfrac{1}{10}B=\dfrac{10^{2011}+1}{10^{2011}+10}=1-\dfrac{9}{10^{2011}+10}\)

10^2012+10>10^2011+10

=>9/10^2012+10<9/10^2011+10

=>-9/10^2012+10>-9/10^2011+10

=>A>B

2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

A = \(\dfrac{n^9+1}{n^{10}+1}\) 

\(\dfrac{1}{A}\) = \(\dfrac{n^{10}+1}{n^9+1}\) = n -  \(\dfrac{n-1}{n^9+1}\)

B = \(\dfrac{n^8+1}{n^9+1}\)

\(\dfrac{1}{B}\) = \(\dfrac{n^9+1}{n^8+1}\) =  n - \(\dfrac{n-1}{n^8+1}\)

Vì n > 1 ⇒ n - 1> 0

       \(\dfrac{n-1}{n^9+1}\) < \(\dfrac{n-1}{n^8+1}\)

⇒ n - \(\dfrac{n-1}{n^9+1}\) > n - \(\dfrac{n-1}{n^8+1}\)⇒ \(\dfrac{1}{A}>\dfrac{1}{B}\)

⇒ A < B 

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

Lời giải:

a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)

Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$

Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$

Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$

b) Rõ ràng $10^{11}-1< 10^{12}-1$. 

Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$

Áp dụng kết quả phần a:

$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$

Cô ơi cho em hỏi là từ 7h - 9h thứ 2 tuần sau tức ngày 29/3 cô có online không ạ ?

Sửa đề: Chứng mình chia hết 24

Tách: 24=8.3

$A={10}^{2012}+{10}^{2011}+{10}^{2010}+{10}^{2009}+8$

⇒$A=\mathrm{10...08}$$⋮$3 (1)

$A=\mathrm{10...008}⋮$8 (Vì: 008$⋮$8) (2)

Từ (1) và (2) ⇒A$⋮$24 Vì: (3,8)

⇒đpcm

tham khảo

https://olm.vn/hoi-dap/detail/48844794829.html

Giải:

a)Ta có:

C=1957/2007=1957+50-50/2007

                      =2007-50/2007

                      =2007/2007-50/2007

                      =1-50/2007

D=1935/1985=1935+50-50/1985

                      =1985-50/1985

                      =1985/1985-50/1985

                      =1-50/1985

Vì 50/2007<50/1985 nên -50/2007>-50/1985

⇒C>D

b)Ta có:

A=2016²⁰¹⁶+2/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1/2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=1+3/2016²⁰¹⁶-1

Tương tự: B=2016²⁰¹⁶/2016²⁰¹⁶-3

                 B=1+3/2016²⁰¹⁶-3

Vì 2016²⁰¹⁶-1>2016²⁰¹⁶-3 nên 3/2016²⁰¹⁶-1<3/2016²⁰¹⁶-3

⇒A<B

Chúc bạn học tốt!

Làm tiếp:

c)Ta có:

M=10²⁰¹⁸+1/10²⁰¹⁹+1

10M=10.(10²⁰¹⁸+1)/20²⁰¹⁹+1

10M=10²⁰¹⁹+10/10²⁰¹⁹+1

10M=10²⁰¹⁹+1+9/10²⁰¹⁹+1

10M=10²⁰¹⁹+1/10²⁰¹⁹+1 + 9/10²⁰¹⁹+1

10M=1+9/10²⁰¹⁹+1

Tương tự:

N=10²⁰¹⁹+1/10²⁰²⁰+1

10N=1+9/10²⁰²⁰+1

Vì 9/10²⁰¹⁹+1>9/10²⁰²⁰+1 nên 10M>10N

⇒M>N

Chúc bạn học tốt!