3B = 1.2.3 + 2.3.(4-1) + ... +2010.2011.(2012-2009)

3B = 1.2.3 + 2.3.4 - 1.2.3 + ..... +2010.2011.2012-2009.2010.2011

3B = 2010.2011.2012

B = 2010.2011.2012:3

B = 2710908440

3B = 1.2.3 + 2.3.(4-1) + ... +2010.2011.(2012-2009)

3B = 1.2.3 + 2.3.4 - 1.2.3 + ..... +2010.2011.2012-2009.2010.2011

3B = 2010.2011.2012

B = 2010.2011.2012:3

B = 2710908440

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

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

b, (x² - 1)(x² - 4)  < 0

=> x² - 1 và x² - 4 khác dấu

Mà x²- 1 > x² - 4 => x² - 1 dương; x² -4 là số âm

=>  0 < x² < 4

=> x² = 1 (Vì x² là số chính phương)

=> x = 1

Vậy.....

a, M = 1.2 + 2.3 +...+ 99.100

=> 3M = 1.2.3 + 2.3.(4 - 1) +...+ 99.100.(101 - 98)

=> 3M = 1.2.3 + 2.3.4 - 1.2.3 +...+ 99.100.101 - 98.99.100

Triệt tiêu các hiệu bằng 0, ta còn:

3M = 99.100.101

=> 3M =999900

=> M = 333300

3M = 3.(1.2 + 2.3 + 3.4 + ... + 212.213 )

= 1.2.3 + 2.3.3 + 3.4.3 + .... + 212.213.3

= 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2 ) + .... + 212.213(214 - 211)

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 212.213.214 - 211.212.213

= 212.213.214

=> M = 212.213.214/3

3M=1.2.3+2.3.(4-1)+3.4.(5-2)+...+212.213.(214-211)

3M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+212.213.214-211.212.213

3M=212.213.214=9663384

M=9663384/3=3221128
 

sao ket qua nhieu vay 

A = 5/1.2 + 5/2.3 +...+ 5/99.100

 2A = 10/1.2 + 10/2.3 +...+ 10/99.100

2 A = 5/1-5/2+5/2-5/3+5/3-5/4+...+5/99-5/100

2A=5/1-5/100

2A=9/2 => A=9/2:2=9/4

cho 1 đ-ú-n-g nha bạn!!

99/20 đảm bảo

lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll

Kẻ Bí Mật sai rồi !                                                                                      

1) Đặt \(A=1.2+2.3+3.4+....+98.99\)

Ta có:\(3A=3.\left(1.2+2.3+3.4+....+98.99\right)\)

\(3A=1.2.3+2.3.3+3.4.3+....+98.99.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+98.99.\left(100-97\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)

\(3A=98.99.100\Rightarrow A=\frac{98.99.100}{3}=323400\)

Ta có:\(\frac{A.y}{1}=184800\Rightarrow y=184800:323400=\frac{4}{7}\)

2)Đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1428+185,8\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{37.38.39}\)

Tổng quát:\(\frac{2}{\left(a-1\right)a\left(a+1\right)}=\frac{1}{\left(a-1\right)a}-\frac{1}{a\left(a+1\right)}\)

Ta có:

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{37.38.39}\)

\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(2B=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\Rightarrow B=\frac{370}{741}:2=\frac{185}{741}\)

Khi đó \(A=\frac{185}{741}.1428+185,8=...........\) (tự tính ra)

(*)số ko đẹp mấy

Đặt biểu thức trên = A

Có : 3A = 1.2.3+2.3.3+3.4.3+....+1001.1002.3

= 1.2.3+2.3.(4-1)+3.4.(5-2)+....+1001.1002.(1003-3)

= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+1001.1002.1003-3.1001.1002

= 1001.1002.1003

=> A = 1002.1002.1003/3 = 335337002

=> A có chữ số tận cùng là 2

k mk nha

\(\frac{31}{20}\)đáp số đúng 1000000000000000000000000000000000000000000000000000% đó nha

h and kb nhé

31/20

31/20 là đúng  đó

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)