1/100.99 - 1/99.98 -.....- 1/2.1
giải giùm mìh nha
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\(C=\frac{1}{100}-\frac{1}{100\times99}-\frac{1}{99\times98}-\frac{1}{98\times97}-...-\frac{1}{3\times2}-\frac{1}{2\times1}\)
\(=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=\frac{1}{100}-\left(\frac{1}{100}-1\right)=\frac{1}{100}-\frac{1}{100}+1=1\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\dfrac{99}{100}=-\dfrac{49}{50}\)
1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1
= 1/100 - (1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1)
= 1/100 - (1/1.2 + 1/2.3 + ... + 1/97.98 + 1/98.99 + 1/99.100)
= 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100)
= 1/100 - (1 - 1/100)
= 1/100 - 99/100
= -49/50
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=-\left(1-\frac{1}{100}\right)\)
\(=-\frac{99}{100}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{2.1}\)
=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)
=\(-\left(1-\frac{1}{100}\right)\)
=\(\frac{-99}{100}\)
A=1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
A= - (1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
A= - (1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
A= - (1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
A= - (1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
A= - (1/1-1/100)
A= - 99/100
à mình nhầm có phải thế này không
1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
=-(1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
=-(1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
=-(1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
=-(1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
=-(1/1-1/100)=-99/100
C= \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
= \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\) ( viet nguoc lai cho de nhin)
= \(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
= \(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
= \(-\frac{49}{50}\)
C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - .... - 1/3.2 - 1/2.1
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{2.1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+...+1-\frac{1}{2}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)=-\frac{1}{2}\)
C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta có A = 1 - 1/100 => C = 1/100 - ( 1 - 1/100) = 1/50 - 1 = -49/50