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\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x=\dfrac{9}{12}-\dfrac{2}{12}\)
\(x=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)
\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)
\(x=\dfrac{1009}{2020}\)
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)
\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)
\(\frac{7}{4}-y\cdot\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\frac{7}{4}-y\cdot\frac{5}{6}=\frac{5}{6}\)
\(y\cdot\frac{5}{6}=\frac{7}{4}-\frac{5}{6}\)
\(y\cdot\frac{5}{6}=\frac{11}{12}\)
\(y=\frac{11}{12}:\frac{5}{6}\)
\(y=\frac{11}{10}\)
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.
\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)
Câu 2:
a. $x\times \frac{2}{7}=\frac{6}{11}$
$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$
b. $x: \frac{3}{2}=\frac{1}{4}$
$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
mình ko viết lại đầu bài nhé
= 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
= 1 - 1/6 = 5/6
trong phép tính đầu mỗi số hạng mk tách làm 1 hiệu nhé
\(1\frac{1}{3}+1\frac{1}{5}.y-\frac{4}{5}=2\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{4}{5}+\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{8}{5}=\frac{18}{5}\)
\(\Rightarrow\frac{6}{5}y=\frac{18}{5}-\frac{4}{3}\)
\(\Rightarrow\frac{6}{5}y=\frac{34}{15}\)
\(\Rightarrow y=\frac{34}{15}:\frac{6}{5}\)
\(\Rightarrow y=\frac{34}{15}.\frac{5}{6}=\frac{17}{9}\)
[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 \(\dfrac{53}{6}\)
= \(\dfrac{159}{10}\)
[5\(\dfrac{1}{3}\) - 2\(\dfrac{2}{3}\)]: \(\dfrac{3}{7}\)
= [\(\dfrac{16}{3}\) - \(\dfrac{8}{3}\)]: \(\dfrac{3}{7}\)
= \(\dfrac{8}{3}\) : \(\dfrac{3}{7}\)
= \(\dfrac{56}{9}\)
y x 3/4 - 1/2 = 2 + 1/5
y x 3/4 - 1/2 = 11/5
y x 3/4 = 11/5 + 1/2
y x 3/4 = 22/10 + 5/10
y x 3/4 = 27/10
y = 27/10 : 3/4
y = 27/10 x 4/3
y = 18/5
y = 18/5