tính bằng cách thuận tiện:
1 và 1/2012 * 1 và 1/2013 * 1 và 1/2014* 1 và 1/2015 :1/2012
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( 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
( 2014 x 2015 - 2016 ) : ( 2012 + 2013 x 2014 )
= ( 4058210 - 2016 ) : ( 2012 + 4054182 )
= 4056194 : 4056194
= 1
\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
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
(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
đó là phép tính của lớp 1 đó hả