1 (3y - 0,8 ) : y + 14,5 = 15

   ( 3y - 0,8 ) : y = 0,5

   3y : y - 0,8 : y = 0,5

   3 - 0,8 : y = 0,5 

   0,8 : y = 2,5

    y = 0,8 : 2,5

    y = 0,32

Ta có :

Tử số = 2012 x 14 + 1997 + 2010 x 2011 

          = ( 2011 + 1 ) x 14 + 1997 + 2010 x 2011

          = 2011 x 14 + 1 x 14 + 1997 + 2010 x 2011

          = 2011 x 14 + 14 + 1997 + 2010 x 2011

          = ( 2011 x 14 ) + ( 14 + 1997 ) + ( 2010 x 2011 )

          = 2011 x 14 + 2011 + 2010 x 2011

          = 2011 x ( 14 + 1 + 2010 )

          = 2011 x 2025

Mẫu số = 2011 x 5 + 2011 x 1008 + 1012 x 2011

            = 2011 x ( 5 + 1008 + 1012 )

            = 2011 x 2025

=> \(A=\frac{2011\times2025}{2011\times2025}=1\)

a. 1⋅2⋅3+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12

= 6+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12

= 6+48+3⋅6⋅9+4⋅8⋅12

= 6+48+162+4⋅8⋅12

= 6+48+162+384

= 600

b . Ta có \(A=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}.\)

Ta có : \(\frac{2010}{2011+2012}< \frac{2010}{2011}\) và \(\frac{2011}{2011+2012}< \frac{2011}{2012}\)

=> \(\frac{2010+2011}{2011+2012}< \frac{2010}{2011}+\frac{2011}{2012}\)

=> A < B

a, 2010 x 3+ 2010 x 6 + 2010

= 2010 x ( 3 + 6 + 1)

= 2010 x 10

= 20100

b, 2011 x 89 + 10 x 2011 + 2011

= 2011 x (89 + 10 + 1)

= 2011 x 100

= 201100

a, 2011 x 3+ 2011 x 6 + 2011

= 2011 x ( 3+6+1)

= 2011 x 10

= 20110

b, 2010 x 89 + 10 x 2010 + 2010

= 2010 x (89+10+1)

= 2010 x 100

= 201000

a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)

\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)

\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)

\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)    (1)

Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)

Nên biểu thức (1) xảy ra khi \(x+2013=0\)

\(x=-2013\)

b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)  (2)

Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)

Nên biểu thức (2) xảy ra khi \(x-2011=0\)

\(x=2011\)

Ta có: x=2011 \(\Rightarrow\)x+1=2012

\(\Rightarrow A=x^{2011}-\left(x+1\right).x^{2010}\)\(+\left(x+1\right)x^{2009}\)\(-\left(x+1\right)x^{2008}+...\)\(-\left(x+1\right)x^2+\left(x+1\right)x-1\)

=\(x^{2011}\)\(-x^{2011}-x^{2010}+x^{2010}+x^{2009}-x^{2009}-\)...\(-x^2+x^2+x-1\)

= \(x-1=2011-1=2010\)

=

Thay 2012=x+1.

\(A=x^{2011}-\left(x+1\right)x^{2010}+\left(x+1\right)x^{2009}-\left(x+1\right)x^{2008}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(A=x^{2011}-x^{2011}-x^{2010}+x^{2010}+x^{2009}-...-x^3-x^2+x^2+x-1\)

\(A=x-1=2011-1=2010\)

mình giải ra trc mà, k mik chứ