D
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
14 tháng 3 2017
=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)
=1+\(\frac{1}{101}\)
=\(\frac{102}{101}\)
14 tháng 3 2017
1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]
1/2.3.4 = 1/2[ 1/2- 1/3 ]
...................
1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]
=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]
A = 1/2 . [1/1.2 -1/100 .101]
A= 1/2 . 5049 /10100 = 5049 / 20200.
Mình nghĩ là vậy đó.
HT
2 tháng 9 2015
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+..+\frac{1}{99.100.101}\)
A = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{101-99}{99.100.101}\right)\)
A = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{101}{99.100.101}-\frac{99}{99.100.101}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100.101}\right)\)
A = \(\frac{1}{2}.\frac{5049}{10100}\)
A = \(\frac{5049}{20200}\)
8 tháng 11 2016
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
=> A = \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
= \(\frac{1}{2}.\frac{5049}{10100}\)
= \(\frac{5049}{20200}\)
8 tháng 11 2016
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\)
Ta thấy:
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{99.100.101}=\frac{1}{99.100}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{10100}\)
\(\Rightarrow2A=\frac{5050}{10100}-\frac{1}{10100}\)
\(\Rightarrow2A=\frac{5049}{10100}\Rightarrow A=\frac{5049}{10100}:2=\frac{5049}{20200}\)
S
31 tháng 8 2017
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
17 tháng 10 2016
\(\frac{1}{2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{99.100.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10100}\right)\)
\(=\frac{1}{2}.\frac{5049}{10100}=\frac{5049}{20200}\)
17 tháng 10 2016
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{5049}{10100}=\frac{5049}{20200}\)
N
2
A
7 tháng 9 2017
Ta có:
\(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+98.99.100.4\)
\(\Rightarrow4A=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...98.99.100.101-97.98.99.100\)
\(\Rightarrow4A=98.99.100.101\)
\(\Rightarrow A=\dfrac{98.99.100.101}{4}\)
Vậy \(A=\dfrac{98.99.100.101}{4}\)
C
7 tháng 9 2017
Ta có: \(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)
\(4A=\left(1.2.3+2.3.4+...+98.99.100\right)4\)
\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)...+98.99.100.\left(101-97\right)\)
\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)
\(4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+97.98.99.100-97.98.99.100+98.99.100.101\)
\(4A=98.99.100.101\)
\(\Rightarrow A=\dfrac{98.99.100.101}{4}=24497550\)
7 tháng 9 2017
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101\
4A=98.99.100.101
A=\(\dfrac{\text{98.99.100.101}}{4}\)
tick nha
C
7 tháng 9 2017
Ta có: \(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)
\(4A=\left(1.2.3+2.3.4+...+98.99.100\right)4\)
\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)...+98.99.100.\left(101-97\right)\)
\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)
\(4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+97.98.99.100-97.98.99.100+98.99.100.101\)
\(4A=98.99.100.101\)
\(\Rightarrow A=\dfrac{98.99.100.101}{4}=24497550\)
TN
1
9 tháng 3 2015
Đặt A=1/1.2.3+1/2.3.4+...+1/99.100.101
2A=2/1.2.3+2/2.3.4+...2/99.100.101
2A=3-1/1.2.3+4-2/2.3.4+...+101-99/99.100.101
2A=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+101/99.100.101-99/99.100.101
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/99.100-1/100.101
2A=1/2-1/10100
DN
5
mình ko biết
A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 99.100.101
4A = 4.(1.2.3 + 2.3.4 + 3.4.5 + ... + 99.100.101)
= 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 99.100.101.(102-98)
= 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + 3.4.5.6 - 3.4.5.6 + ... + 98.99.100.101 - 98.99.100.101 + 99.100.101.102
4A = 99.100.101.102
A = 99.100.101.102 : 4
A = 25497450