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\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
\(\left(1-\frac{1}{2018}\right)\times\left(1-\frac{1}{2019}\right)\times\left(1-\frac{1}{2020}\right)\times\left(1-\frac{1}{2021}\right)\times\left(1-\frac{1}{2022}\right)\)
\(=\frac{2017}{2018}\times\frac{2018}{2019}\times\frac{2019}{2020}\times\frac{2020}{2021}\times\frac{2021}{2022}\)
\(=\frac{2017}{2022}\)
x +\(\frac{1}{6}\)=\(\frac{1}{3}\)-\(\frac{1}{4}\)
x +\(\frac{1}{6}\)=\(\frac{1}{12}\)
x =\(\frac{1}{12}\)-\(\frac{1}{6}\)
x =\(-\frac{1}{12}\)
x -\(\frac{5}{12}\)=\(\frac{1}{3}+\frac{1}{4}\)
x -\(\frac{5}{12}\)=\(\frac{7}{12}\)
x =\(\frac{7}{12}\)+\(\frac{5}{12}\)
x = 1
x +\(\frac{1}{2}\)=\(\frac{4}{5}\)
x =\(\frac{4}{5}\)-\(\frac{1}{2}\)
x =\(\frac{3}{10}\)
x -\(\frac{1}{6}\)=\(\frac{1}{2}\)+\(\frac{1}{3}\)
x -\(\frac{1}{6}\)=\(\frac{5}{6}\)
x =\(\frac{5}{6}\)+\(\frac{1}{6}\)
x = 1
x +\(\frac{1}{8}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)
x +\(\frac{1}{8}\)=\(\frac{1}{6}\)
x =\(\frac{1}{6}\)-\(\frac{1}{8}\)
x =\(\frac{1}{24}\)
Học tốt!!!
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neu the minh ket ban kieu gi
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\)= \(\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}\)= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\)= \(\frac{1.2.3}{2.3.4}=\frac{1}{4}\)
\(\left(1+\frac{1}{2}\right):\left(1+\frac{1}{3}\right):\left(1+\frac{1}{4}\right)\)= \(\frac{2+1}{2}:\frac{3+1}{3}:\frac{4+1}{4}\)= \(\frac{3}{2}:\frac{4}{3}:\frac{5}{4}=\frac{3}{2}.\frac{3}{4}.\frac{4}{5}=\frac{3.3.4}{2.4.5}=\frac{3.3}{2.5}=\frac{9}{10}\)
\(\frac{6}{5}-\frac{4}{5}.\frac{4}{8}+1=\frac{6}{5}+1-\frac{4.4}{5.8}=\frac{6+5}{5}-\frac{4}{5.2}=\frac{11}{5}-\frac{2}{5}=\frac{9}{5}\)
x < \(\frac{3}{4}+\frac{4}{5}=\frac{15}{20}+\frac{16}{20}=\frac{31}{20}=\frac{20}{20}+\frac{11}{20}=1\frac{11}{20}\)=> x = 0 ; 1
\(\frac{1}{2}+\frac{2}{3}< x< 4-\frac{5}{4}\)<=> \(\frac{3}{6}+\frac{4}{6}< x< \frac{16}{4}-\frac{5}{4}\)<=> \(\frac{7}{6}< x< \frac{11}{4}\)=> \(\frac{14}{12}< \frac{12x}{12}< \frac{33}{12}\)
=> 14 < 12x < 33 => 12x = 24 => x = 24 : 12 = 2
Bài 1
1/2 x 3/4 : 4/5
= 3/8 : 4/5
= 3/8 x 5/4
= 15/32
2/3 + 1/6 - 1/2
= 4/6 + 1/6 - 3/6
= 5/6 - 3/6
= 1/3
3/5 + 4 : 2/3 - 3/2
= 3/5 + 4 x 3/2 - 3/2
= 3/5 + 6 - 3/2
= 3/5 + 30/5 - 3/2
= 33/5 - 3/2
= 66/10 - 15/10
= 28/5
Bài 2
x - 1/4 = 1/2
x = 1/2 + 1/4
x = 2/4 + 1/4
x = 3/4
2/3 + x = 1
x = 1 - 2/3
x = 3/3 - 2/3
x = 1/3
(2/3 x 4/5) x 5/6
= 8/15 x 5/6
= 4/9
(2/3 x 4/5) x 5/6
= (5/6 x 4/5) x 2/3
= 2/3 x 2/3
= 4/9
(1/2 + 1/3) x 1/5
= 5/6 x 1/5
= 1/6
(1/2 + 1/3) x 1/5
= (1/2 x 1/5) + (1/3 x 1/5)
= 1/10 + 1/15
= 1/6
4A6?