d) 0

( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 1 + 1/3 - 4/3)

=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 4/3 - 4/3)

=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x 0

=0

Ta có: \(\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(\dfrac{3}{3}+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

=0

\(A=\frac{2013^{2014}+1}{2013^{2015}+1}\)

\(\Rightarrow2013A=\frac{2013\left(2013^{2014}+1\right)}{2013^{2015}+1}=\frac{2013^{2015}+2013}{2013^{2015}+1}\)(1)

\(B=\frac{2013^{2012}+1}{2013^{2013}+1}\)

\(\Rightarrow2013B=\frac{2013\left(2013^{2012}+1\right)}{2013^{2013}+1}=\frac{2013^{2013}+2013}{2013^{2013}+1}\)(2)

Từ (1) và (2) => A<B

Hinh nhu minh hoc ca hai bang nhau ma

Dãy số có 2 chữ số chia hết cho 3 là:[12,15,....,99] 

Khoảng cách của từng số hạng là 3

Số số hạng là: (99-12):3+1=30(số)

Vậy có 30 số có 2 chữ số chia hết cho 3

( 2014 x 2015 - 2016 ) : ( 2012 + 2013 x 2014 )

= ( 4058210 - 2016 ) : ( 2012 + 4054182 )

= 4056194 : 4056194

= 1

﴾ 2014 x 2015 ‐ 2016 ﴿ : ﴾ 2012 + 2013 x 2014 ﴿

= ﴾ 4058210 ‐ 2016 ﴿ : ﴾ 2012 + 4054182 ﴿

= 4056194 : 4056194

= 1

 ( 2014 x 2015 - 2016 ) : (2012 + 2013 x 2014 )

= ( 2014 x ( 2013 + 2 ) - 2016 ) : ( 2012 + 2013 x 2014 )

= ( 2014 x 2013 + 2014 x 2 - 2016 ) : ( 2012 + 2013 x 2014 )

= ( 2014 x 2 - 2016 ) : 2012

=  ( 4028 - 2016 ) : 2012 

=      2012       : 2012

= 1 

\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)

=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)

1/ Ta có : \(\frac{212121}{353535}=\frac{3.7.10101}{5.7.10101}=\frac{3}{5}=\frac{42}{70}\)

\(\frac{131313}{141414}=\frac{13.10101}{14.10101}=\frac{13}{14}=\frac{65}{70}\)

Vì 42 < 65 =>  \(\frac{42}{70}<\frac{65}{70}\)  =>  \(\frac{212121}{353535}<\frac{131313}{141414}\)

Ta có P(x)= x⁴+ax³+bx²+cx+d

Đặt P(x)= (x-2013)(x-2014)(x-2015)(x-x₀)+mx²+nx+p

P(2013)=2014=>4052169m+2013n+p=2014} m=0

P(2014)=2015=>4056196m+2014n+p=2015}=> n=1

P(2015)=2016=>4060225m+2015n+p=2016} p=1

=>P(x)= (x-2013)(x-2014)(x-2015)(x-x₀)+x+1

=>.) P(2012)= -6(2012-x₀)+2012+1

= -12072+6x₀+2013=-10059+6x₀

.)P(2016)=6(2016-x₀)+2016+1

=12096-6x₀+2017=14113-6x₀

=> P(2012)+P(2016)= -10059+6x₀+14113-6x₀=4054

1)Ta có : 212121/353535 = 212121:10101/353535:10101 = 21/35 = 3/5

131313/141414 = 131313:10101/141414:10101 = 13/14

Ta có : 3/5 = 42/70 ; 13/14 = 65/70

Vì 42<65 => 42/70<65/70 => 212121/353535<131313/141414

2)Ta có : 2012/2013<1 

2013/2014<1

2014/2015<1

=> 2012/2013+ 2013/2014+ 2015/2012<1+1+1=3

Vậy 2012/2013+ 2013/2014+ 2015/2012<3

Ai chơi truy kích cấp cao thì kết bạn nha !!!!!!!!

212121/353535=21/35=3/5

131313/141414=13/14. 

b, 2012/2013<1; 2013/2014<1; 2014/2015<1 nên  2012/2013+2013/2014+ 2014/2015<3

(2011/2012+2012/2013+2013/2014+...+3026/3027) x (1/5-2/3:3/10)

= (2011/2012+2012/2013+2013/2014+...+3026/3027) x (1/5-2/10)

= (2011/2012+2012/2013+2013/2014+...+3026/3027) x (1/5-1/5)

= (2011/2012+2012/2013+2013/2014+...+3026/3027) x 0

= 0

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thanks

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