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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)
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.

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 = \(\frac{5049}{20200}\)

19 tháng 6 2018

\(A=\frac{5049}{20200}\)

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:

\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{99.100.101}=\frac{1}{99.100}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{10100}\)

\(\Rightarrow2A=\frac{5050}{10100}-\frac{1}{10100}\)

\(\Rightarrow2A=\frac{5049}{10100}\Rightarrow A=\frac{5049}{10100}:2=\frac{5049}{20200}\)

 

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

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)

\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

\(\Leftrightarrow A=\frac{1}{2}.\frac{5049}{10100}=\frac{5049}{20200}\)

A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)

=1/2*(1/2-1/99*100)

=1/2*(4950-1/9900)

=4950/19800

14 tháng 4 2019

\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)

\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)

19 tháng 7 2016

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(=>2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{98.99.100}\)

Dễ dàng CM đẳng thức phụ sau : \(\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\)

Áp dụng vào tính 2B,ta có:

\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+....+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}=>B=\frac{4949}{9900}:2=\frac{4949}{19800}\)

Vậy.....

19 tháng 7 2016

1/1.2.3 + 1/2.3.4 + .... + 1/98.99.100

= 1/2(1/1.2-1/2.3) + 1/2(1/2.3-1/3.4) + ..... + 1/2(1/98.99-1/99.100)

= 1/2(1/1.2-1/2.3+1/2.3-....+1/98.99-1/99.100)

= 1/2(1/2 - 1/9900)

= 1/2(4950/9900 - 1/9900)

= 1/2. 4949/9900

= 4949/19800

10 tháng 2 2017

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10100}\right)=\frac{5049}{20200}\)

10 tháng 2 2017

A=1/2(1/1.2-1/2.3+1/2.3-1/3.4+.......+1/99.100-1/100.101)

A=1/2(1/1.2-1/100.101)

A=1/2(1/1.2-1/100.101)=5049/202000

10 tháng 2 2017

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{5049}{20200}\)

10 tháng 2 2017

1/1.2.3+1/2.3.4+...+1/99.100.101

= 1/2 ( 1/1.2-1/2.3+1/2.3-1/3.4+...+1/99.100-1/100.101)

=1/2(1/1.2-1/.100.101)=5049/20200

10 tháng 2 2017

\(\frac{1}{1.2.3}+\frac{1}{1.2.3}+...+\frac{1}{99.100.101}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10100}\right)=\frac{5049}{20200}\)