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AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Câu 1:
a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$
b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)
c.
\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)
d.
\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Câu 2:
a. $x\times \frac{2}{7}=\frac{6}{11}$
$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$
b. $x: \frac{3}{2}=\frac{1}{4}$
$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 10 2023
[4\(\dfrac{1}{5}\) - 2\(\dfrac{2}{5}\)] x 8\(\dfrac{5}{6}\)
= [\(\dfrac{21}{5}\) - \(\dfrac{12}{5}\)] x \(\dfrac{53}{6}\)
= \(\dfrac{9}{5}\) x \(\dfrac{53}{6}\)
= \(\dfrac{159}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 10 2023
[5\(\dfrac{1}{3}\) - 2\(\dfrac{2}{3}\)]: \(\dfrac{3}{7}\)
= [\(\dfrac{16}{3}\) - \(\dfrac{8}{3}\)]: \(\dfrac{3}{7}\)
= \(\dfrac{8}{3}\) : \(\dfrac{3}{7}\)
= \(\dfrac{56}{9}\)
25 tháng 4 2021
\(\dfrac{2}{5}\cdot x=\dfrac{4}{7}:\dfrac{2}{3}\)
\(\dfrac{2}{5}\cdot x=\dfrac{6}{7}\)
\(x=\dfrac{6}{7}:\dfrac{2}{5}\)
\(x=\dfrac{15}{7}\)
PT
25 tháng 4 2021
\(\dfrac{2}{5}.x=\dfrac{4}{7}:\dfrac{2}{3}\)
\(\dfrac{2}{5}.x=\dfrac{4}{7}.\dfrac{3}{2}\)
\(\dfrac{2}{5}.x=\dfrac{6}{7}\)
\(x=\dfrac{6}{7}:\dfrac{2}{5}\)
\(x=\dfrac{6}{7}.\dfrac{5}{2}=\dfrac{15}{7}\)
4 tháng 12 2021
\(\frac{2}{3}\)x \(\frac{5}{8}\)x \(\frac{8}{15}\)= \(\frac{2}{3}\)x \(\frac{1}{15}\)= \(\frac{2}{45}\)
\(\frac{22}{5}\)x \(12\)x \(\frac{20}{40}\)= \(\frac{22}{5}\)x \(12\)x \(\frac{1}{2}\)= \(\frac{22}{5}\)x 6 = \(\frac{122}{5}\)
\(\frac{7}{2}\)x \(\frac{26}{7}\)x \(\frac{4}{13}\)= \(\frac{91}{7}\)x \(\frac{4}{13}\)
NH
2
14 tháng 2 2022
a: =>x=5/6-3/4=10/12-9/12=1/12
b: =>x=12-3/5=57/5
c: =>x=3/5+5/7=21/35+25/35=46/35
d: =>x=7/9:3=7/27
e: =>x=7/8x12/5=84/40=21/10
f: =>x=13:5/3=39/5
BD
15 tháng 6 2023
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x=\dfrac{9}{12}-\dfrac{2}{12}\)
\(x=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)
\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)
\(x=\dfrac{1009}{2020}\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 6 2023
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)
\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)
\(\frac{4}{7}\cdot\frac{2}{3}:x=\frac{4}{5}\)
(=)\(\frac{2}{3}x=\frac{4}{5}:\frac{4}{7}\)
(=)\(\frac{2}{3}x=\frac{7}{5}\)
(=) \(x=\frac{7}{5}:\frac{2}{3}\)
(=) \(x=\frac{21}{10}\)
4/7 . 2/3 : x = 4/5
8/21 : x = 4/5
x = 8/21 : 4/5
x = 10/21