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\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
Biến đổi tử số
\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)
= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)
= \(\dfrac{20}{20}+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{1}{19}\)
= 20 x \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)\)
Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
= \(\dfrac{20\times\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}=20\)
Vậy A = 20
ta có :
`1/5=(1xx8)/(5xx8)=8/40`
`5/8=(5xx5)/(8xx5)=25/40`
`7/20=(7xx2)/(20xx2)=14/40`
`19/40=19/40`
`->5/8` lớn nhất
`=>B`
\(\dfrac{1}{10}+\dfrac{17}{20}+\dfrac{3}{10}+\dfrac{2}{20}+\dfrac{6}{10}+\dfrac{1}{20}\)
= \(\left(\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{6}{10}\right)+\left(\dfrac{17}{20}+\dfrac{2}{20}+\dfrac{1}{20}\right)\)
= 1 + 1 = 2
= ( 1/10 + 3/10 + 6/10 ) + ( 17/20 + 2/20 + 1/20 ) = 1+1 = 2
1/10
x = \(\dfrac{2}{20}=\dfrac{1}{10}\)