tinh gia tri cua bieu thuc A=1.2+2.3+3.4+...2005.2006
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{31}{20}\)đáp số đúng 1000000000000000000000000000000000000000000000000000% đó nha
h and kb nhé
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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: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)