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a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
a)x*4/5:2=6/7 b)6/5:x:5/4=10/15
x*4/5=6/7 x 2 6/5:x=10/15 x 5/4
x*4/5=12/7 6/5:x=5/6
x=12/7:4/5 x=6/5:5/6
x=15/7 x=36/25
a) x * 4/5 : 2 = 6/7
x * 4/5 = 6/7 *2
x *4/5 = 12/7
x = 12/7 : 4/5
x = 15/7
Vậy x = 15/7
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
Ta có:
\(A=10\times\left(1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\right)\)
\(=10\times\left[9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\right]\)
\(=10\times\left[9-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\right]\)
\(=10\times\left[9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\right]\)
\(=10\times\left[9-\left(1-\frac{1}{10}\right)\right]\)
\(=10\times\left(9-\frac{9}{10}\right)\)
\(=10\times\frac{81}{10}\)
\(=81\)
Vậy; A = 81
=.= hk tốt!!
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
Bài 1:
A = 1996 x 1997 x 1998 x 1999 + 2021 x 2022 x 2023 x 2024
A = (1996 x 1997) x (1998 x 1999) + (2021 x 2022) x (2023 x 2024)
A = \(\overline{..2}\) x \(\overline{..2}\) + \(\overline{..2}\) x \(\overline{..2}\)
A = \(\overline{..4}\) + \(\overline{..4}\)
A = \(\overline{..8}\)
\(\left(\dfrac{5}{6}+\dfrac{7}{x}\right)-\dfrac{16}{12}=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{5}{6}+\dfrac{7}{x}=\dfrac{2}{3}+\dfrac{16}{12}=\dfrac{8}{12}+\dfrac{16}{12}=\dfrac{24}{12}=2\)
\(\Rightarrow\dfrac{7}{x}=2-\dfrac{5}{6}=\dfrac{12}{6}-\dfrac{5}{6}=\dfrac{7}{6}\)
=> x = 6
(\(\dfrac{5}{6}\) + \(\dfrac{7}{x}\)) - \(\dfrac{16}{12}\) = \(\dfrac{2}{3}\)
\(\dfrac{5}{6}\) + \(\dfrac{7}{x}\) = \(\dfrac{2}{3}\) + \(\dfrac{16}{12}\)
\(\dfrac{5}{6}\) + \(\dfrac{7}{x}\) = 2
\(\dfrac{7}{x}\) = 2 - \(\dfrac{5}{6}\)
\(\dfrac{7}{x}\) = \(\dfrac{1}{6}\)
\(x\) = 6 x (7 : 1)
\(x\) = 6 x 7
\(x\) = 42
#Yu