\(\frac{2011.2012-1}{2001.2012}\)\(=\frac{2012.2013-1}{2012.2013}\)vì rút gọn hai phân số ta đều được kết quả là \(\frac{-1}{1}\)

Ta có \(\frac{2011.2012-1}{2011.2012}\)=\(\frac{2011^2+2011-1}{2011^2+2011}\)

Lại có \(\frac{2012.2013-1}{2012.2013}\)=\(\frac{2012^2+2012-1}{2012^2+2012}\)

Mặt khác có \(\frac{a}{b}\)<\(\frac{a+m}{b+m}\)(với a<b )

\(\Rightarrow\)\(\frac{2012^2+2012-1}{2012^2+2012}\)<...............

còn lại bn tự làm nha dễ lắm

Ta có:              \(\dfrac{2012.2013-1}{2012.2013}\)=\(\dfrac{2012.2013}{2012.2013}-\dfrac{1}{2012.2013}\)(1)

                        \(\dfrac{2011.2012-1}{2011.2012}\)=\(\dfrac{2011.2012}{2011.2012}-\dfrac{1}{2011.2012}\)(2)

                                  Ta thấy:\(\dfrac{1}{2011.2012}>\dfrac{1}{2012.2013}\)=>(1)>(2)  (vì số bị trừ đều như nhau mà số trừ lớn hơn nên b/thức đó bé hơn)

                         Vậy \(\dfrac{2011.2012-1}{2011.2012}< \dfrac{2012.2013-1}{2012.2013}\)

a)N=\(\frac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^6}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}=\frac{2^{30}\cdot3^{18}\cdot5-2^{29}\cdot3^{20}}{2^{28}\cdot3^{18}\cdot\left(5\cdot3-7\cdot2\right)}\)

\(=\frac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-3\cdot3\right)}{2^{28}\cdot3^{18}}=\frac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}=2\)

Vậy N=2

choáng thế

\(A=\left(1-\frac{1}{2010}\right)-\left(1-\frac{1}{2011}\right)+\left(1-\frac{1}{2012}\right)-\left(1-\frac{1}{2013}\right)=-\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)

\(A=-\frac{1}{2010.2011}-\frac{1}{2012.2013}\)

Vì 2010.2011 > 2009.2010 => \(\frac{1}{2010.2011}-\frac{1}{2009.2010}\)

\(-\frac{1}{2012.2013}>-\frac{1}{2011.2012}\)

=> A > B

alna marian: Công thức gì vậy bạn?

S = 1/2 - 1/3 + 1/3 -1/4 + ......... +1/2011 -1/2012

S= 1/2 - 1/2012 = 1005/2012

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{2012}\)

\(S=\frac{1}{2}+0+0+0+...-\frac{1}{2012}\)

\(S=\frac{1}{2}-\frac{1}{2012}\)

\(S=\frac{1005}{2012}\)

\(A=\frac{2012}{1}\cdot\frac{1005}{2012}\)

\(A=1005\)

\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)

\(\frac{1}{2013}x+1+(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013})=2\)

\(\frac{1}{2013}x+1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+1+\left(1-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+1+1-\frac{1}{2013}=2\)

\(\frac{1}{2013}x-\frac{1}{2013}+2=2\)

\(\frac{1}{2013}.\left(x-1\right)=2-2\)

\(\frac{1}{2013}.\left(x-1\right)=0\)

=> x - 1 = 0

x = 1

\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)

\(\frac{1}{2013}x+\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\)

\(\frac{1}{2013}x+\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+\left(1-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+\frac{2012}{2013}=2\)

\(\frac{1}{2013}x=2-\frac{2012}{2013}\)

\(\frac{1}{2013}x=\frac{2014}{2013}\)

\(x=\frac{2014}{2013}:\frac{1}{2013}\)

=> x=2014

a, m = 19.90 = 19.3.30 = 57.30

    n = 31.60 = 31.2.30 = 62.30

    n > m

b, 2011 < 2015

    2012 < 2015

  2011.2012 < 2015.2015

   p < q