mk bít lm cách lớp 5, vừa học

Cần ko bn

Làm bậy, mà đúng

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)

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

= \(\frac{1}{1}\)- \(\frac{1}{100}\)

= \(\frac{99}{100}\)

    1/1 . 2 + 1/ 2 . 3 + 1/ 3 . 4 + ... + 1/99 . 100

= 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100

= 1/1 - 1/100 

= 100/100 + -1/100

= 99/100

            #Hoq chắc _ Baccanngon

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Câu hỏi của Vũ Thị Kim Oanh - Toán lớp 7 - Học toán với OnlineMath

= 1/1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +.....+1/99 + 1/100

=( 1/1 + 1/2 +1/3 +1/4 + 1/5 + 1/6 +.....1/99 + 1/100) - 2(1/2 + 1/4 + 1/6 + .....+ 1/100)

=(1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 +.....+ 1/99 + 1/100) - ( 1 + 1/2 + 1/3 + .... + 1/50)

= 1/51 + 1/52 + 1/53 +....+ 1/100....>1/100

= ( 1/51 + 1/52 + 1/53 +.....+ 1/75) + ( 1/76 + 1/77 + 1/78 +.....+ 1/100)

Có 1/51>1/52>1/53>....>1/75 ; 1/76>1/77>1/78>....>1/100

A> 1/75.25 + 1/100.25= 1/3 + 1/4 = 7/12

A< 1/51.25+ 1/76.25 < 1/50.25 + 1/75.25= 1/2+1/3=5/6

Vậy 7/12< A< 5/6

a) 1 + 2 + 3 + 4 + ... + 100

= (100 + 1) x 100 : 2

= 5050

a) A=(100-1):1+1=100 số hạng   

    A=100:2=50 cặp

    tính giá trị của từng cặp số = (1+100)+(2+99)+(3+98)+...+(50+51)=101

    tính giá trị của biểu thức A: 50*101=5050

    [ mình tính theo công thức đó ]

\(y=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=\left(1+\frac{1}{2}+\frac{1}{3}...+\frac{1}{100}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

=>y=B

=>y-B=0

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

vay ket =\(\frac{99}{100}\)

con lai tu tinh nhe

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

=1/2-1/100=49/100

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2.\left(1-\frac{1}{99}\right)\)

\(=2.\frac{98}{99}\)

\(=\frac{196}{99}=1\frac{97}{99}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(\frac{1}{1}-\frac{1}{100}\)

=>\(\frac{99}{100}\)

B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)

=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)

=>\(\frac{1}{1}-\frac{1}{99}\)

=>\(\frac{98}{99}\)