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\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
Ta có: \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)=\frac{2.19}{20}=\frac{19}{10}\)
\(\frac{2}{1\times2}+\frac{2}{2\times3}+......+\frac{2}{19\times20}\)
\(=2\left(\frac{1}{1\times2}+\frac{1}{2\times3}+.......+\frac{1}{19\times20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)=2.\frac{19}{20}=\frac{19}{10}\)
\(2\times\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{18x19}+\frac{1}{19x20}\right)\)
\(2x\left(1-\frac{1}{20}\right)=2x\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=1.\left(1-\frac{1}{20}\right)\)
\(=1.\frac{19}{20}\)
\(=\frac{19}{20}\)
Lưu ý: Từ bước thứ 2 bạn chuyển thành số La-tinh nhé.
P/s: "." là nhân nhé.
B = \(\dfrac{2}{1\times2}\) + \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+ \(\dfrac{2}{99\times100}\)
B = 2 \(\times\) ( \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+....+ \(\dfrac{1}{99\times100}\))
B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+...+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\))
B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))
B = 2 \(\times\) \(\dfrac{99}{100}\)
B = \(\dfrac{99}{50}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Cho hai số biết rằng bớt số thứ nhất 28 đơn vị thì được số thứ hai va 1/3 số thứ nhất bằng 3/5 số thứ hai.Tìm hai số đó
ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
=> \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)
a ) 3/5 + 3/16 + 13/16
= 3/5 + ( 3/16 + 13/16 )
= 3/5 + 16/16
= 3/5 + 1
= 3/5 + 5/5
= 8/5
b ) 1/1 x 2 + 1/ 2 x 3 + 1/ 3 x 4
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4
= 1 - 1/4
= 4/4 - 1/4
= 3/4
\(\frac{1}{1x2}+\frac{1}{1x3}+...+\frac{1}{999x1000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)
1/1x2+1/2x3+1/3x4+...+1/999x1000
=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000
=1-1/1000
=1000/1000-1/1000
=999/1000
M=1x2+2x3+3x4+xxx+19x20
3M=1x2x3+2x3x3+3x4x3+xxx+19x20x3
3M=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+xxx+19x20x(21-18)
3M=(1x2x3-0x1x2)+(2x3x4-1x2x3)+(3x4x5-2x3x4)+xxx+(19x20x21-18x19x20)
3M=19x20x21-0x1x2
3M=7980-0
3M=7980
M=7980:3
M=2660
Vậy M=2660
Chúc bạn học tốt nha!!!!! ^^
\(A=1\cdot2+2\cdot3+...+19\cdot20\)
\(=1\left(1+1\right)+2\left(2+1\right)+...+19\left(19+1\right)\)
\(=\left(1^2+2^2+...+19^2\right)+\left(1+2+...+19\right)\)
\(=\dfrac{19\left(19+1\right)\left(2\cdot19+1\right)}{6}+\dfrac{19\cdot20}{2}\)
\(=2470+190=2660\)