1.

\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{2012}{2013}\)

\(A=\frac{1.2.3.4.....2012}{2.3.4.5......2013}\)

\(A=\frac{1}{2013}\)

\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)

\(B=\frac{2012\left(2013-2012\right)}{2012\left(2011+2\right)}\)

\(B=\frac{2013-2012}{2011+2}\)

\(B=\frac{1}{2013}\)

\(Vì:\frac{ 1}{2013}=\frac{1}{2013}\)

\(\Rightarrow\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)

\(Hay: A=B\)

\(A=\frac{1\times2}{2\times2}\times\frac{2\times3}{3\times3}\times\frac{3\times4}{4\times4}\times\frac{4\times5}{5\times5}\times...\times\frac{2012\times2013}{2013\times2013}\)

\(\Rightarrow A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2012}{2013}\)

\(\Rightarrow A=\frac{1\times2\times3\times4\times...\times2012}{2\times3\times4\times5\times...\times2013}\)

\(\Rightarrow A=\frac{1}{2013}\)

\(B=\frac{2012\times2013-2012\times2012}{2012\times2011+2012\times2}\)

\(\Rightarrow B=\frac{2012\times\left(2013-2012\right)}{2012\times\left(2011+2\right)}\)

\(\Rightarrow B=\frac{2012\times1}{2012\times2013}\)

\(\Rightarrow B=\frac{1}{2013}\)

A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+....+98.99(100-97)   "." la dau nhan

A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+98.99.100-97.98.99

A=1.2.3+98.99.100

A= 970206

Ta có : B = 1.2 + 2.3 + 3.4 + ..... + 98.99

=> 3B = 0.1.2 + 1.2.3 - 1.2.3 + ...... + 98.99.100

=> 3B = 98.99.100

=> B = \(\frac{98.99.100}{3}\)  = 323400

C = 1.1 + 2.2 + 3.3 + ..... + 98.98

=> C = 1.(2 - 1) + 2(3 - 1) + 3(4 - 1) + ...... + 98(99 - 1)

=> C = 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ..... + 98.99 - 98

=> C = (1.2 + 2.3 + 3.4 + ..... + 98.99) - (1 + 2 + 3 + ..... + 98) 

=> C = 323400 - 4851

=> C = 318549. 

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2}-0+0+...+0-\frac{1}{100}\)

\(\Rightarrow\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)

Ta thấy:

1/2*2<1/1*2)vì 2*2>1*2).

1/3*3<1/2*3(vì 3*3>2*3).

...

1/8*8<1/7*8(vì 8*8>7*8).

=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.

=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.

=>B<1-1/8.

=>B<7/8.

Mà 7/8<1.

=>B<1.

Vậy B<1(đpcm).

\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(\Rightarrow1-\frac{1}{8}< 1\)

=>B<1

\(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6.}+\frac{1}{7.7}+\frac{1}{8.8}\)\(=\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(B=1-\frac{1}{8}\)

\(\Rightarrow B< 1\left(ĐPCM\right)\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow M=1-\frac{1}{100}\)

\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)

\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)

\(a,M=1-\frac{1}{100}=\frac{99}{100}\)

\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)

                  \(=1-\frac{1}{99}=\frac{98}{99}\)

   =>\(N=\frac{98}{99}:2=\frac{49}{99}\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=1-\frac{1}{100}\)

\(M=\frac{99}{100}\)

a)(y+2):5-5x5=378

(y+2):5-25=378

(y+2):5=378+25

(y+2):5=403

(y+2)=403x5

y+2=2015

y=2015-2

y=2013

(y+2):5-5.5=378

(y+2):5-25=378

(y+20)=378+25

(y+2)=403

(y+2)=403.5

y+2=2015

y=2015-2

y=2013

a) \(\left(y+2\right):5-5.5=378\)

\(\left(y+2\right):5-25=378\)

\(\left(y+2\right):5=378+25\)

\(\left(y+2\right):5=403\)

\(y+2=403.5\)

\(y+2=2015\)

\(y=2015-2\)

\(y=2013\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)

\(1-\frac{1}{x+1}=\frac{499}{500}\)

\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)

=> x + 1 = 500

=> x = 500 - 1

=> x = 499

Vậy x = 499

1/1.2 + 1/2.3 + 1/3.4 +...+ 1/x.(x+1)=499/500

1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 +...+ 1/x -1/(x+1) =499/500

1-1/(x+1)=499/500

=>x/(x+1)=499/500

=>x=499

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(\Rightarrow5A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

\(\Rightarrow5A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\right)\)

\(\Rightarrow5A=1-\frac{1}{8}\)

\(\Rightarrow A=\left(1-\frac{1}{8}\right).\frac{1}{5}=\frac{7}{40}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{5}{7.8}\right)\)

\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=5\left(1-\frac{1}{8}\right)\)

\(A=5.\frac{7}{8}\)

\(A=\frac{38}{8}\)

nhầm rồi :)

\(A=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{8}\right)\)

\(\Rightarrow A=\frac{5.7}{8}\)

\(\Rightarrow A=\frac{35}{8}\)