Mình làm mẫu 1 bài nha !

Có : 12A = 1.5.12+5.9.12+....+101.105.12

= 1.5.12+5.9.(13-1)+.....+101.105.(109-97)

= 1.5.12+5.9.13-1.5.9+.....+101.105.109-97.101.105

= 1.5.12-1.5.9+101.105.109

= 1155960

=> A = 1155960 : 12 = 96330

Tk mk nha

Có : 4D = 1.2.3.4+2.3.4.4+....+98.99.100.4

= 1.2.3.4+2.3.4.(5-1)+.....+98.99.100.(101-97)

= 1.2.3.4+2.3.4.5-1.2.3.4+......+98.99.100.101-97.98.99.100

= 98.99.100.101

=> D = 98.99.100.101/4 = 24497550

mk k vt lại đề nha

S=2.(1/1.2+1/2.3+1/3.4+............+1/99.100)

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

S=2.(1-1/100)

S=2.99/100

S=198/100

S=\(\frac{2}{1.2}\)+\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)+...+\(\frac{2}{98.99}\)+\(\frac{2}{99.100}\)

S=\(\frac{2}{1}\).(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{98.99}\)+\(\frac{1}{99.100}\))

S=\(\frac{2}{1}\).(\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{98}\)-\(\frac{1}{99}\)+\(\frac{1}{99}\)-\(\frac{1}{100}\))

S=\(\frac{2}{1}\).(\(\frac{1}{1}\)-\(\frac{1}{100}\))

S=\(\frac{2}{1}\).(\(\frac{100}{100}\)-\(\frac{1}{100}\))

S=\(\frac{2}{1}\).\(\frac{99}{100}\)

S=\(\frac{99}{50}\)

Vậy S=\(\frac{99}{50}\)

\(P=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\\ =2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\\ =2\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\cdot\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =2\cdot\dfrac{99}{100}\\ =\dfrac{99}{50}\)

\(P=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\\ =2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\\ =2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)

\(P=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(1-\dfrac{1}{100}\right)\)
\(=2\cdot\dfrac{99}{100}\)
\(=\dfrac{99}{50}\)
#DatNe

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

= \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

= \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

 =\(2.\frac{99}{100}\)

 =\(\frac{99}{50}\)

bạn hãy rút gọn vế phải: x/200=1/2.2/3.3/4......98/99.99/100

  Rồi sẽ có cái phương trình:x/200=1/100

từ đó suy ra:x/200=2/200 =>x=2

:)))))

\(\dfrac{x}{200}=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}\)

\(\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{100}\)

\(\Leftrightarrow x=2\)

1. ta có :

\(3^2+4^2=5^{x-1}\)

  \(25=5^{x-1}\)

 \(5^2=5^{x-1}\)

=> x = 3

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101

=> 3S = 99.100.101

=> S = 99.100.101/3

=> S = 333300