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a: =6x5/4=30/4=15/2
b: =8x7/12=56/12=14/3
c: =5x8/7=40/7
d: =15x3/5=9
a: =6x5/4=30/4=15/2
b: =8x7/12=56/12=14/3
c: =5x8/7=40/7
d: =15x3/5=9
a: =25/30+12/30=37/30
b: =24/20-15/20=9/20
c: =36/48=3/4
d: =8/17x1/6=8/102=4/51
\(\dfrac{1}{2}-\dfrac{3}{8}=\dfrac{4}{2\times4}-\dfrac{3}{8}=\dfrac{4}{8}-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\dfrac{4}{3}-\dfrac{8}{15}=\dfrac{4\times5}{3\times5}-\dfrac{8}{15}=\dfrac{20}{15}-\dfrac{8}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\dfrac{5}{6}-\dfrac{7}{12}=\dfrac{5\times2}{6\times2}-\dfrac{7}{12}=\dfrac{10}{12}-\dfrac{7}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)
\(\dfrac{11}{4}-\dfrac{9}{8}=\dfrac{11\times2}{4\times2}-\dfrac{9}{8}=\dfrac{22}{8}-\dfrac{9}{8}=\dfrac{13}{8}\)
\(\dfrac{17}{16}-\dfrac{3}{4}=\dfrac{17}{16}-\dfrac{3\times4}{4\times4}=\dfrac{17}{16}-\dfrac{12}{16}=\dfrac{5}{16}\)
\(\dfrac{31}{36}-\dfrac{5}{6}=\dfrac{31}{36}-\dfrac{5\times6}{6\times6}=\dfrac{31}{36}-\dfrac{30}{36}=\dfrac{1}{36}\)
`6/11+2/5+16/11+19/13+3/5+7/13`
`=(6/11+16/11)+(2/5+3/5)+(19/13+7/13)`
`=22/11+5/5+26/13`
`=2+1+2=5`
a: =4/5+1/5+2/3+1/3=1+1=2
b: =17/12+7/12+29/7-8/7=3+2=5
c: =3/5+2/5+16/7-1/7-1/7
=1+2=3
d: =2/5+3/5+2/3+1/3+7/4+1/4
=1+1+2
=4
a) \(\dfrac{16}{15}+\dfrac{7}{15}+\dfrac{4}{15}=\left(\dfrac{16}{15}+\dfrac{4}{15}\right)+\dfrac{7}{15}=\dfrac{20}{15}+\dfrac{7}{15}=\dfrac{27}{15}\)
b) \(\dfrac{5}{17}+\dfrac{7}{17}+\dfrac{13}{17}=\dfrac{5}{17}+\left(\dfrac{7}{17}+\dfrac{13}{17}\right)=\dfrac{5}{17}+\dfrac{20}{17}=\dfrac{25}{17}\)
\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
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
☹ huhu
\(\dfrac{16}{17}-\dfrac{2}{3}=\dfrac{48}{51}-\dfrac{34}{51}=\dfrac{14}{51}\)