A= 1\(\dfrac{1}{2}\) \(\times\)1\(\dfrac{1}{3}\) \(\times\)1\(\dfrac{1}{4}\) \(\times\)1\(\dfrac{1}{5}\) \(\times\).... \(\times\)1\(\dfrac{1}{2020}\) \(\times\) 1\(\dfrac{1}{2021}\)
a.\(\dfrac{2}{3}\times\dfrac{1}{4}-\dfrac{1}{3}\times\dfrac{1}{2}\) =
b.\(\dfrac{8}{5}\times\dfrac{1}{4}-\dfrac{2}{5}\times\dfrac{1}{2}-\dfrac{1}{2}\times\dfrac{1}{5}=\)
giải rõ ràng cho mình nhé
a) \(\dfrac{2}{3}\times\dfrac{1}{4}-\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{2}{12}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=\dfrac{0}{6}=0\)
b) \(\dfrac{8}{5}\times\dfrac{1}{4}-\dfrac{2}{5}\times\dfrac{1}{2}-\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{8}{20}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4-2-1}{10}=\dfrac{1}{10}\)
a. 1/6 - 1/6 = 0 (hoặc 2/12 - 1/6= 2/12 - 2/12 = 0)
b. 4/5 - 1/2 x ( 2/5 - 1/5 ) = 4/5 - 1/2 x 1/5
= 4/5 x 2/10
= 4/25
>; <; =?
a) \(\dfrac{2}{3}\times\dfrac{4}{5}\) \(\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}\) \(\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}+\dfrac{2}{15}\right)\times\dfrac{3}{4}\) \(\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
Tính:
a) \(\dfrac{2}{5}\times\dfrac{3}{8}\times\dfrac{3}{4}\) b) \(\dfrac{1}{3}\times\dfrac{1}{6}\times\dfrac{1}{9}\)
c) \(\dfrac{3}{4}:\dfrac{1}{5}:\dfrac{7}{8}\) d) \(\dfrac{3}{5}:\dfrac{1}{5}:\dfrac{1}{8}\)
a) $\frac{2}{5} \times \frac{3}{8} \times \frac{3}{4} = \frac{{2 \times 3 \times 3}}{{5 \times 8 \times 4}} = \frac{{18}}{{160}} = \frac{9}{{80}}$
b) $\frac{1}{3} \times \frac{1}{6} \times \frac{1}{9} = \frac{{1 \times 1 \times 1}}{{3 \times 6 \times 9}} = \frac{1}{{162}}$
c) $\frac{3}{4}:\frac{1}{5}:\frac{7}{8} = \frac{3}{4} \times \frac{5}{1} \times \frac{8}{7} = \frac{{3 \times 5 \times 8}}{{4 \times 1 \times 7}} = \frac{{120}}{{28}} = \frac{{30}}{7}$
d) $\frac{3}{5}:\frac{1}{5}:\frac{3}{8} = \frac{3}{5} \times \frac{5}{1} \times \frac{8}{3} = \frac{{3 \times 5 \times 8}}{{5 \times 1 \times 3}} = 8$
Tìm A biết: A\(\times\)(1-\(\dfrac{1}{4}\))\(\times\)(1-\(\dfrac{1}{9}\))\(\times\)(1-\(\dfrac{1}{16}\))\(\times\)(1-\(\dfrac{1}{25}\))=1\(\dfrac{3}{5}\)
\(A\cdot\left(1-\dfrac{1}{4}\right)\cdot\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)=1\dfrac{3}{5}\)
=>\(A\cdot\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{5}\right)=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{5}\cdot\dfrac{6}{2}=\dfrac{8}{5}\)
=>\(A\cdot3=8\)
=>A=8/3
tính A= \((1-\dfrac{1}{1+2})\times(1-\dfrac{1}{1+2+3})\times(1-\dfrac{1}{1+2+3+4})\times...\times(1-\dfrac{1}{1+2+3+4+5+...+n})\)
a) Tính rồi so sánh:
\(\dfrac{1}{2}\times\dfrac{1}{3}\) \(\dfrac{1}{3}\times\dfrac{1}{2}\) \(\dfrac{3}{5}\times\dfrac{1}{6}\) \(\dfrac{1}{6}\times\dfrac{3}{5}\)
Nhận xét: Khi thực hiện phép nhân hai phân số, ta có thể đổi chỗ các phân số trong một tích mà tích của chúng không thay đổi.
b) Lấy ví dụ tương tự câu a rồi đố bạn thực hiện
a) $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ ; $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$
Vậy $\frac{1}{2} \times \frac{1}{3} = \frac{1}{3} \times \frac{1}{2}$
$\frac{3}{5} \times \frac{1}{6} = \frac{3}{{30}} = \frac{1}{{10}}$ ; $\frac{1}{6} \times \frac{3}{5} = \frac{3}{{30}} = \frac{1}{{10}}$
Vậy $\frac{3}{5} \times \frac{1}{6} = \frac{1}{6} \times \frac{3}{5}$
b) Học sinh tự thực hiện
Tính :
\(\dfrac{1}{2}\times\dfrac{2}{3}\times.........\times\dfrac{2019}{2020}\)
\(=\dfrac{1\cdot2\cdot...\cdot2019}{2\cdot3\cdot...\cdot2020}=\dfrac{1}{2020}\)
A = \(\dfrac{-19}{9}\times\dfrac{1}{2}-\dfrac{4}{11}\times\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)\)
B = \(\left(-\dfrac{15}{6}\right)\div\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}\times\dfrac{-11}{2}\)
C = \(\dfrac{3}{4}\times\left(-8\right)-\dfrac{1}{3}\times\dfrac{-7}{2}-\dfrac{5}{18}\)
\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)
\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)
\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)
\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)
\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)
Tính.
a) \(\dfrac{4}{7}\times\dfrac{3}{5}\) b) \(\dfrac{1}{3}\times\dfrac{9}{10}\) c) \(\dfrac{1}{6}\times\dfrac{1}{4}\)
a) \(\dfrac{4}{7}\times\dfrac{3}{5}=\dfrac{4\times3}{7\times5}=\dfrac{12}{35}\)
b) \(\dfrac{1}{3}\times\dfrac{9}{10}=\dfrac{1\times9}{3\times10}=\dfrac{9}{30}=\dfrac{3}{10}\)
c) \(\dfrac{1}{6}\times\dfrac{1}{4}=\dfrac{1}{4\times6}=\dfrac{1}{24}\)