Đặt a=2013

\(\Rightarrow M=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\frac{a^2+2a+1+a^4+2a^3+a^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\frac{\left(a^4+2a^3+a^2\right)+2\left(a^2+a\right)+1}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\left(\frac{a^2+a+1}{a+1}\right)^2}+\frac{a}{a+1}\)

\(\Rightarrow M=\frac{a^2+a+1+a}{a+1}\)(Bỏ trị tuyệt đối vì a=2013)

\(\Rightarrow M=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=1013+1=1014\)

\(\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1015}{1014}\right)\)

\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1014}{1014}\right).\left(1-\frac{1015}{1014}\right)\)

\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-1\right).\left(1-\frac{1015}{1014}\right)\)

\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...0.\left(1-\frac{1015}{1014}\right)\)

\(=0\)

`A=(10^14-1)/(10^15-11)`

`=>10A=(10^15-10)/(10^15-11)`

`=>10A=(10^15-11+1)/(10^15-11)`

`=>10A=1+1/(10^15-1)`

`=>A>1/10`

`B=(10^14+1)/(10^15+9)`

`=>10B=(10^15+10)/(10^15+9)`

`=>10A=(10^15+9+1)/(10^15+9)`

`=>10A=1+1/(10^15+9)`

Vì `1/(10^15-1)>1/(10^15+9)`

`=>10B>10A`

`=>B>A`

Giải:

\(A=\dfrac{10^{14}-1}{10^{15}-11}\) 

\(10A=\dfrac{10^{15}-10}{10^{15}-11}\) 

\(10A=\dfrac{10^{15}-11+1}{10^{15}-11}\) 

\(10A=1+\dfrac{1}{10^{15}-11}\) 

Tương tự:

\(B=\dfrac{10^{14}+1}{10^{15}+9}\) 

\(10B=\dfrac{10^{15}+10}{10^{15}+9}\) 

\(10B=\dfrac{10^{15}+9+1}{10^{15}+9}\) 

\(10B=1+\dfrac{1}{10^{15}+9}\) 

Vì \(\dfrac{1}{10^{15}-11}>\dfrac{1}{10^{15}+9}\) nên \(10A>10B\) 

\(\Rightarrow A>B\) 

Chúc bạn học tốt!

ta có: 1-(1014/1015)= 1/1015
         1-(2014/2015)= 1/2015
vì 1/1015>1/2015 =>1014/1015<2014/2015
VẬY 1014/1015<2014/2015

có : 1-1014/1015=1/1015

1-2014/2015=1/2015

do 1/1015>1/2015

suy ra 1014/1015<2014/2015

1014/1015<2014/2015

Lời giải:

\(A-6=5^1+5^2+...+5^{2015}\)

\(5(A-6)=5^2+5^3+...+5^{2016}\)

Trừ theo vế:
\(4(A-6)=5^{2016}-5^1\)

\(\Rightarrow A=\frac{5^{2016}-5}{4}+6=\frac{5^{2016}+19}{4}\)

--------------

\(B=\frac{5^{1015}(5^{1001}+2)-10.5^{1014}-1}{4}=\frac{5^{2016}+2.5^{1015}-2.5^{1015}-1}{4}\)

\(=\frac{5^{2016}-1}{4}< \frac{5^{2016}+19}{4}\)

Do đó \(B< A\)

C=1009.1015=1009.1013+2.1009

=1011.1013-2.1013+2.1009

=1011.1013-8

Do 1011.1013-8>1011.1013-512

=>A>B

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