Tính giá trị của biểu thức:
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
Giải giùm nhé! Mình cảm ơn nhìu nhìu! ^__^
\(\frac{1\times2}{2\times3}+\frac{2\times3}{3\times4}+\frac{3\times4}{4\times5}+...+\frac{98\times99}{99\times100}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
Tính biểu thức A
\(A=\frac{5}{1\times2}+\frac{5}{2\times3}+\frac{5}{3\times4}+...+\frac{5}{98\times99}+\frac{5}{99\times100}\)
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=5/1-5/2+5/2-5/3+...+5/99-5/100
A=5-(5/2-5/2+5/3-5/3+...5/99-5/99)-5/100
A=5-0+0+0+...+0-5/100
A=5-5/100
A=49/10=4,9
Bài 4 : Tính nhanh :
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
Tính \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{98\times99\times100}\)
sud kênh Mik ủng hộ với tên kênh là M.ichibi
kênh làm về MINECRAFT
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
tự tính
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{99\cdot99}-\frac{1}{99\cdot100}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}=\frac{4949}{9900}\Rightarrow A=\frac{4949}{19800}\)
Tinh:
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+......+\frac{2}{98\times99}+\frac{2}{99\times100}\)
kết quả cuối cùng là 198/100
\(\frac{2}{1X2}+\frac{2}{2X3}+\frac{2}{3X4}+...+\frac{2}{98X99}+\frac{2}{99X100}\)
\(2X\left(\cdot\frac{1}{1X2}+\frac{1}{2X3}+...+\frac{1}{98X99}+\frac{1}{99X100}\right)\)
\(2X\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(2X\left(1-\frac{1}{100}\right)\)
\(2X\frac{99}{100}\)
\(\frac{99}{50}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{98.99}+\frac{2}{99.100}\)
\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(2.\left(1-\frac{1}{100}\right)\)
\(2.\frac{99}{100}\)\(=\frac{99}{50}\)
\(A=\left[1-\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+......+\frac{1}{98\times99\times100}\right)\right]\times\frac{14851}{19800}\)
Tính :
A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
B = \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
Câu A
Ta có (1/2)A = 1/22 + 1/23 + ... + 1/2100 + 1/2101
=> (1/2)A - A = - (1/2)A = (1/22 + 1/23 + ... + 1/2100 + 1/2101) - (1/2 + 1/22 + ... + 1/2100 )
= 1/2101 - 1/2
=> A = 1 - 1/2100
Câu B
Ta có 1/(1x2) = 1/1 - 1/2
1/(2.3) = 1/2 - 1/3
.................................
1/(99.100) = 1/99 - 1/100
=> B = 1/1 - 1/2 + 1/2 - 1/3 +.... +1/99 - 1/100
= 1 - 1/100
=99/100
\(y=\frac{1\times100+2\times99+3\times98...+99\times2+100\times1}{1\times2+2\times3+3\times4+...+99\times100+100\times101}=?\)
\(\frac{1}{1\times2}\)+ \(\frac{1}{2\times3}\)+ \(\frac{1}{3\times4}\)+ ....... + \(\frac{1}{98\times99}\)+ \(\frac{1}{99\times100}\)
đang cần gấp !!!!!!!!
Lời giải :
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
ko chép lại đề :
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ......... + \(\frac{1}{98}\)- \(\frac{1}{99}\)+ \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Cbht