Tính:
2x3x5
______
4x9x10
K
Khách
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Những câu hỏi liên quan
D
11 tháng 1 2023
\(\dfrac{2\times3\times5}{4\times9\times10}\)
\(=\dfrac{2\times3\times5}{2\times2\times3\times3\times5\times2}\)
\(=\dfrac{1}{2\times3\times2}=\dfrac{1}{12}\)
NX
2
MN
3 tháng 6 2016
a)\(\frac{7x9x12}{6x14x18}=\frac{7x9x6x2}{6x7x2x9x2}=\frac{2}{2x2}=\frac{1}{2}\)
b) \(\frac{2x3x5}{4x9x10}=\frac{2x3x5}{2x2x3x3x2x5}=\frac{1}{2x2x3x5}=\frac{1}{60}\)
31 tháng 1 2020
1. Rút gọn
\(\frac{8x20x4}{9x20x4}=\frac{8}{9}\)
\(\frac{7x9x12}{6x14x18}=\frac{1x1x2}{1x2x2}=\frac{2}{4}=\frac{1}{2}\)
\(\frac{2x3x5}{4x9x10}=\frac{1x1x1}{2x3x2}=\frac{1}{12}\)
2. tìm x
\(\frac{x}{27}=\frac{12}{18}\)
\(\frac{x}{27}=\frac{2}{3}\)
=> X x 3 = 27 x 2
X x 3 = 54
X = 54 : 3
X = 18
\(\frac{35}{49}=\frac{10}{x}\)
\(\frac{5}{7}=\frac{10}{x}\)
=> 5 x X = 7 x 10
5 x X = 70
X = 70 : 5
X = 14
1 tháng 2 2020
1. \(\frac{8.20.4}{9.20.4}\)= \(\frac{8}{9}\)
\(\frac{7.9.12}{6.14.18}\)= \(\frac{7.9.6.2}{6.7.2.9.2}\)= \(\frac{1}{2}\)
\(\frac{2.3.5}{4.9.10}\)= \(\frac{2.3.5}{2.2.3.3.2.5}\)= \(\frac{1}{12}\)
2. \(\frac{x}{27}\)= \(\frac{12}{18}\) \(\frac{35}{49}\)= \(\frac{10}{x}\)
Ta có : \(\frac{12}{18}\)= \(\frac{6}{9}\) Ta có : \(\frac{35}{49}\)= \(\frac{70}{98}\)
Mà \(\frac{x}{27}\)= \(\frac{6}{9}\) Mà \(\frac{10}{x}\)= \(\frac{70}{98}\)
\(\Rightarrow\)\(x\)= \(27:9.6\) \(\Rightarrow\)\(x\)= \(98:\left(70:10\right)\)
Chú ý : '' \(.\)" là dấu nhân.
NT
3
6 tháng 8 2018
\(\frac{30\cdot25\cdot7\cdot8}{75\cdot8\cdot12\cdot14}\)
\(=\frac{2\cdot3\cdot5\cdot5\cdot5\cdot7\cdot8}{5\cdot3\cdot5\cdot8\cdot3\cdot2\cdot2\cdot2\cdot7}\)
\(=\frac{5}{3\cdot2\cdot2}\)
\(=\frac{5}{12}\)
\(\frac{2\cdot3\cdot5}{20\cdot6}\)
\(=\frac{2\cdot3\cdot5}{4\cdot5\cdot2\cdot3}\)
\(=\frac{1}{4}\)
PM
6 tháng 8 2018
\(\frac{30\times25\times7\times8}{75\times8\times12\times14}=\frac{6\times5\times25\times7\times8}{3\times25\times8\times6\times2\times7\times2}=\frac{5}{3\times2\times2}=\frac{5}{12}\)
\(\frac{2\times3\times5}{20\times6}=\frac{2\times3\times5}{4\times5\times3\times2}=\frac{1}{4}\)
~Học tốt ~~
CD
1
LT
6
15 tháng 9
Đặt \(A=1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103\)
\(=1\cdot2\cdot\left(3+1\right)+2\cdot3\cdot\left(4+1\right)+\cdots+100\cdot101\cdot\left(102+1\right)\)
\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)
Đặt \(B=1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\)
\(=\left(2-1\right)\cdot2\cdot\left(2+1\right)+\left(3-1\right)\cdot3\cdot\left(3+1\right)+\cdots+\left(101-1\right)\cdot101\cdot\left(101+1\right)\)
\(=2\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+101\left(101^2-1\right)\)
\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2+3+\cdots+101\right)\)
\(=\left(1^3+2^3+3^3+\cdots+101^3\right)-1-\left(2+3+\cdots+101\right)\)
\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1+2+3+\cdots+101\right)\)
\(=\left(1+2+3+\cdots+101\right)^2-\left(1+2+3+\cdots+101\right)\)
\(=\left\lbrack101\cdot\frac{102}{2}\right\rbrack^2-101\cdot\frac{102}{2}=\left(101\cdot51\right)^2-101\cdot51\)
Đặt \(C=1\cdot2+2\cdot3+\cdots+100\cdot101\)
\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+100\left(100+1\right)\)
\(=\left(1^2+2^2+\cdots+100^2\right)+\left(1+2+\cdots+100\right)\)
\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{100\cdot101}{2}=\frac{100\cdot101\cdot201}{6}+50\cdot101\)
\(=50\cdot101\cdot67+50\cdot101=50\cdot101\cdot68\)
Ta có: A\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)
=B+C
\(=\left(101\cdot51\right)^2-101\cdot51+50\cdot101\cdot68\)
\(=101^2\cdot51^2-101\cdot51+50\cdot101\cdot68=101\left(101\cdot51^2-51+50\cdot68\right)=101\cdot266050\)
Đặt \(D=1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2\)
\(=2^2\left(2-1\right)+3^2\left(3-1\right)+\cdots+101^2\left(101-1\right)\)
\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2^2+3^2+\cdots+101^2\right)\)
\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1^2+2^2+\cdots+101^2\right)\)
\(=\left(1+2+\cdots+101\right)^2-101\cdot\frac{\left(101+1\right)\left(2\cdot101+1\right)}{6}\)
\(=\left(101\cdot\frac{102}{2}\right)^2-101\cdot17\cdot2023=101^2\cdot51^2-101\cdot17\cdot2023\)
\(=101\cdot17\left(101\cdot17\cdot3^2-2023\right)=101\cdot17\cdot13430\)
Ta có: \(\frac{1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103}{1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2}\)
\(=\frac{101\cdot266050}{101\cdot17\cdot13430}=\frac{1565}{1343}\)
\(\dfrac{2\times3\times5}{4\times9\times10}=\dfrac{2\times3\times5}{2^2\times3^2\times2\times5}=\dfrac{1}{2^2\times3}=\dfrac{1}{12}\)
\(\dfrac{2\text{×}3\text{×}5}{4\text{×}9\text{×}10}=\dfrac{2\text{×}3\text{×}5}{2\text{×}2\text{×}3\text{×}3\text{×}2\text{×}5}=\dfrac{1}{2\text{×}3\text{×}2}=\dfrac{1}{12}\)