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.\(\dfrac{2}{3}\times\dfrac{1}{4}-\dfrac{1}{3}\times\dfrac{1}{2}\) =
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) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
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\(\dfrac{1}{2021}\)
\(A=1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times....\times1\dfrac{1}{2020}\times1\dfrac{1}{2021}\\ =\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times....\times\dfrac{2022}{2021}\\ =\dfrac{3\times4\times5\times6\times.....\times2022}{2\times3\times4\times5\times....\times2021}\\ =\dfrac{2022}{2}=1011\)
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) $\frac{3}{5}:\frac{1}{5}:\frac{3}{8} = \frac{3}{5} \times \frac{5}{1} \times \frac{8}{3} = \frac{{3 \times 5 \times 8}}{{5 \times 1 \times 3}} = 8$
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
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 \(\left(1+\dfrac{1}{1+2}\right)\times\left(1+\dfrac{1}{1+2+3}\right)\times\left(1+\dfrac{1}{1+2+3+4}\right)\times...\times\left(1+\dfrac{1}{1+2+3+...+997}\right)\)
Bài 2: Tìm x:
a)\(\dfrac{x-1}{27}\)=\(\dfrac{-3}{1-x}\) c)\(3\times x=2\times y\) và\(x-2\times y=8\)
b)\(\dfrac{4}{5}\)-\(\left|x-\dfrac{1}{2}\right|\)=\(\dfrac{3}{4}\) d)\(\dfrac{x-1}{2005}\)=\(\dfrac{3-y}{2006}\) và x-4009=y
a: \(\Leftrightarrow\left(x-1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Tính (theo mẫu).
Mẫu: \(5\times\dfrac{2}{9}=\dfrac{5}{1}\times\dfrac{2}{9}=\dfrac{5\times2}{1\times9}=\dfrac{10}{9}\) Ta có thể viết gọn như sau: \(5\times\dfrac{2}{9}=\dfrac{5\times2}{9}=\dfrac{10}{9}\) |
a) \(3\times\dfrac{4}{11}\) b) \(1\times\dfrac{5}{4}\) c) \(0\times\dfrac{2}{5}\)
a) \(3\times\dfrac{4}{11}=\dfrac{3\times4}{11}=\dfrac{12}{11}\)
b) \(1\times\dfrac{5}{4}=\dfrac{1\times5}{4}=\dfrac{5}{4}\)
c) \(0\times\dfrac{2}{5}=\dfrac{0\times2}{5}=\dfrac{0}{5}=0\)
a: \(=\dfrac{3\cdot4}{11}=\dfrac{12}{11}\)
b: \(=\dfrac{1\cdot5}{4}=\dfrac{5}{4}\)
c: \(=\dfrac{0\cdot2}{5}=0\)