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BU
6 tháng 4 2022
\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
AD
23 tháng 7 2023
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
23 tháng 7 2023
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
5 tháng 11 2021
\(\Leftrightarrow\dfrac{15}{16}:x=1-\dfrac{1}{12}=\dfrac{11}{12}\)
\(\Leftrightarrow x=\dfrac{15}{16}:\dfrac{11}{12}=\dfrac{15}{16}\cdot\dfrac{12}{11}=\dfrac{45}{44}\)
25 tháng 8 2023
a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)
\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)
\(3-a=\dfrac{3}{2}\)
\(a=3-\dfrac{3}{2}\)
\(a=\dfrac{6}{2}-\dfrac{3}{2}\)
\(a=\dfrac{3}{2}\)
b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)
\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)
\(0-a=\dfrac{7}{8}\)
\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)
c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)
\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)
\(\dfrac{8}{6}-a=\dfrac{1}{6}\)
\(a=\dfrac{8}{6}-\dfrac{1}{6}\)
\(a=\dfrac{7}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)
a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)
a = 2 + 1 - 1 - \(\dfrac{1}{2}\)
a = 2 - \(\dfrac{1}{2}\)
a = \(\dfrac{3}{2}\)
b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
(3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)
a = - \(\dfrac{7}{8}\)
c, 2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a = \(\dfrac{1}{6}\)
a = 2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)
a = (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)
a = 1 + \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)
a = \(\dfrac{7}{6}\)
17 tháng 1 2023
(y - \(\dfrac{1}{2}\)) : \(\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\right)\)= \(\dfrac{1}{3}\)
(y\(-\dfrac{1}{2}\)): \(\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)= \(\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\dfrac{3}{10}=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right)=\dfrac{1}{10}\)
y = \(\dfrac{3}{5}\)
\(1-\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\)
\(=1-\dfrac{1}{1+\dfrac{1}{\dfrac{3}{2}}}\)
\(=1-\dfrac{1}{1+\dfrac{2}{3}}\)
\(=1-\dfrac{1}{\dfrac{5}{3}}\)
\(=1-\dfrac{3}{5}\)
\(=\dfrac{2}{5}\)