Tinh gia tri bieu thuc A = \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
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Ta thấy: 1/1x2= 1/1-1/2
1/2x3= 1/2-1/3...
1/99x100= 1/99-1/100
Vậy A= 1-1/2+1/2-1/3+...1/99- 1/100= 1-1/100= 99/100
( Thông cảm vì máy tính của mình không có phần mềm để biểu thị phân số nên đành viết gạch chéo vậy)
Ta có :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
TA CÓ: \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}\)= \(\frac{99}{100}\)
Ta có : \(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}=\frac{99}{100}\)
Bài này lớp 6 phải không bạn
A=1/1-1/2+1/2-1/3+1/3-1/4+1/5-1/6+......................+1/99-1/100
A=1/1-1/100
A=99/100
Nếu bạn cảm thấy bài mình đúng thì cho mình một "lai"
1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100
=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100
=(1-1/100)+(1/2 - 1/2 ) + ( 1/3 - 1/3 ) + ...... + (1/98 - 1/98 ) + ( 1/99 - 1/99 )
= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0
=99/100
vậy GTBT = 99/100
A = 5(1/1.2 + 1/2.3 +......+ 1/99.100)
A = 5( 1 - 1/2 + 1/2 - 1/3 +........+ 1/99 - 1/100)
A = 5( 1 - 1/100)
A = 5 . 99/100
A = 99/20
** k mk nha!
\(\frac{5}{1\times2}+\frac{5}{2\times3}+...+\frac{5}{99\times100}=5\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\right)=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5\times\frac{99}{100}=\frac{99}{20}=4\frac{19}{20}\)
\(A=\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\Rightarrow A=5\left(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{5x99}{100}=\frac{99}{20}\)
\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+\frac{5}{3}-\frac{5}{4}+....+\frac{5}{99}-\frac{5}{100}\)
\(A=\frac{5}{1}+\left(-\frac{5}{2}+\frac{5}{2}\right)+\left(-\frac{5}{3}+\frac{5}{3}\right)+\left(-\frac{5}{4}+\frac{5}{4}\right)+...\left(-\frac{5}{99}+\frac{5}{99}\right)+\frac{5}{100}\)
\(A=\frac{5}{1}+0+0+....+0+\frac{5}{100}\)
\(A=\frac{500}{100}+\frac{5}{100}=\frac{205}{100}=\frac{101}{20}\)
Đúng 100%
Đúng 100%
Đúng 100%
A = 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ... + 1/99x 100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100
A =1 - 1/100
A = 99/100
\(\frac{1}{1\times2}+\frac{1}{2\times3}+...........+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)