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

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

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

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

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

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

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

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

anh chiu

chán thế

3.4-9.2+x=1.2

9.2+x=3.4-1.2

9.2+x=2.2

x=2.2-9.2

x=-7

3,4-9,2+x=1,2

     9,2+x=3,4-1,2

     9,2+x=2,2

           x=2,2-9,2

           x= -7

\(C=\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.8}+\frac{1}{9.10}\right)\)

\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{10}\right)\)

\(C=\frac{9}{10}-\frac{9}{10}=0\)

bài A: áp dụng công thức: 1 + 2 + 3 + ... + n = n x (n + 1) : 2 tính được 5050

bài B: áp dụng công thức:  \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)  rồi triệt tiêu gần hết, qui đồng mẫu số tính được B = 99/100

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

    = ( 100 + 1 ) x 100 : 2 = 5050

Vậy A = 5050

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

   \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

   \(=1-\frac{1}{100}\)

   \(=\frac{99}{100}\)

Vậy \(B=\frac{99}{100}\) 

Học tốt #

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)

\(\left(1-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)

\(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)

\(\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)

\(\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\Rightarrow x+\frac{266}{100}=5\Rightarrow x=\frac{117}{50}\)

Vậy x = 117/50

Ta có:

 \(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right).100\\ =\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100\)

   \(=\left(1-\frac{1}{10}\right).100\)    

   \(=\frac{9}{10}.100\)

   = 90

Khi đó đề bài sẽ thành : \(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)

                                \(\Rightarrow\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)

                                \(\Rightarrow\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\)

                                \(\Rightarrow x+\frac{266}{100}=5\)

                               \(\Rightarrow x=\frac{117}{50}\)

Vậy \(x=\frac{117}{50}\)

A=1.2+2.3+3.4+4.5+...+99.100

=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300

học sinh lớp 5 trở suống ko làm đc đâu

Ta có A=\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

            =\(\frac{1.2+2.2.2+3.3.2+4.4.2+5.5.2}{3.4+3.2.4.2+3.3.4.3+3.4.4.4+3.5.4.5}\)

            =\(\frac{2.\left(4+9+16+25\right)}{3.4.\left(4+9+16+25\right)}\)

            =\(\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)

           B=\(\frac{111111}{666666}=\frac{1}{6}\)

=>A=B

l-ike mình nha

\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot2\left(1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10\right)}=\frac{1}{6}\)

\(B=\frac{111111}{666666}=\frac{1\cdot111111}{6\cdot111111}=\frac{1}{6}\)

Vì 1/6 =1/6 nên A=B

A = \(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

A = \(\frac{1.2.\left(1+2+3+4+5\right)}{3.4.\left(1+2+3+4+5\right)}\)

A = \(\frac{2}{12}=\frac{222222}{1333332}\)

B = \(\frac{111111}{666665}=\frac{222222}{1333330}\)

Vì \(\frac{222222}{1333332}<\frac{222222}{1333330}\)

=> A < B

\(=\frac{14\cdot101+15\cdot101+...+19\cdot101}{20\cdot101+21\cdot101+...+25\cdot101}=\frac{101\cdot\left(14+15+16+17+18+19\right)}{101\cdot\left(20+21+22+23+24+25\right)}\)

\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}=\frac{\left(14+19\right)+\left(15+18\right)+\left(16+17\right)}{\left(20+25\right)+\left(21+24\right)+\left(22+23\right)}=\frac{33.3}{45.3}=\frac{33}{45}=\frac{11}{15}\)

\(=\frac{2015}{2016}\)

2015/2016 nhé bạn.