a = 2014¹⁰⁰ - 2014⁹⁹ = 2014⁹⁹(2014 - 1) = 2014⁹⁹.2013

b = 2014⁹⁹ - 2014⁹⁸ = 2014⁹⁸(2014 - 1) = 2014⁹⁸.2013

Vì 2014⁹⁹.2013 > 2014⁹⁸.2013 => a > b

A= 2014^100 - 2014^99 = 2014^99 ( 2014 -1) = 2014^99 . 2013 

B = 2014^99 - 2014^98 = 2014^98 ( 2014 - 1) = 2013.2014^98 

Vì 2014^98 <2014^99 > 2013.2014^98 < 2013.2014^99 

=> B < A

kb đc 0

2 câu đầu tôi làm đc

Ta có: \(M=1\cdot2014+2\cdot2013+\cdots+2014\cdot1\)

\(=2\left(1\times2014+2\times2013+\cdots+1007\times1008\right)\)

\(=2\left\lbrack1\times\left(2015-1\right)+2\times\left(2015-2\right)+\cdots+1007\times\left(2015-1007\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times\left(1+2+\cdots+1007\right)-\left(1^2+2^2+\cdots+1007^2\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times\frac{1008}{2}-\frac{1007\times\left(1007+1\right)\times\left(2\times1007+1\right)}{6}\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times504-1007\times168\times2015\right\rbrack=2\times2015\times1007\times168\left(3-1\right)=4\times168\times2015\times1007\)

\(N=1+\left(1+2\right)+\cdots+\left(1+2+\cdots+2014\right)\)

\(\) \(=\frac{1\times2}{2}+\frac{2\times3}{2}+\cdots+\frac{2014\times2015}{2}\)

\(=\frac12\times\left(1\times2+2\times3+\cdots+2014\times2015\right)\)

\(=\frac12\times\left\lbrack1\times\left(1+1\right)+2\times\left(2+1\right)+\cdots+2014\times\left(2014+1\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\left(1\times1+2\times2+\cdots+2014\times2014\right)+\left(1+2+\cdots+2014\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\frac{2014\times\left(2014+1\right)\times\left(2\times2014+1\right)}{6}+\frac{2014\times2015}{2}\right\rbrack\)

\(=\frac12\times\left\lbrack1007\times2015\times1343+1007\times2015\right\rbrack=\frac12\times1007\times2015\times\left(1343+1\right)\)

=1007x2015x672

=4x168x2015x1007

Do đó: M=N

A lớn hơn

M < N

Đáp án 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 \(A< \frac{1}{4}\)

Ta có công thức : 

\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)

\(A=\frac{99^{2015}+1}{99^{2014}+1}>\frac{99^{2015}+1+98}{99^{2014}+1+98}=\frac{99^{2015}+99}{99^{2014}+99}=\frac{99\left(99^{2014}+1\right)}{99\left(99^{2013}+1\right)}=\frac{99^{2014}+1}{99^{2013}+1}=B\)

\(\Rightarrow\)\(A>B\)

Chúc bạn học tốt ~ 

Ôi ***** :)) bạn thêm vào cho mình mấy từ ạ :<< cop xg mà nó mất chữ :((
Dòng thứ nhất : Ta có : A = ...

Dòng mà B = .... thêm vào : Lại có B = ....

Dòng gần cuối : Như vậy ta có A/B = ....

Ta có:

\(\frac{2013}{2014}>\frac{2013}{2014+2015}\)

\(\frac{2014}{2015}>\frac{2014}{2014+2015}\)

\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\)

\(\Rightarrow M>N\)

Ta có: \(N=\frac{2013+2014}{2014+2015}<1\);

          \(M=\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013}{2015}+\frac{2014}{2015}=\frac{4027}{2015}>1\)

\(\Rightarrow A>B\)