11./17÷33/34
K
Khách
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DM
6
MN
28 tháng 5 2021
\(\dfrac{7}{9}\cdot\dfrac{3}{35}=\dfrac{7}{3\cdot3}\cdot\dfrac{3}{7\cdot5}=\dfrac{1}{15}\)
\(\dfrac{9}{22}\cdot55=\dfrac{9}{11\cdot2}\cdot11\cdot5=\dfrac{45}{2}\)
\(\dfrac{11}{17}:\dfrac{33}{34}=\dfrac{11}{17}\cdot\dfrac{34}{33}=\dfrac{11\cdot17\cdot2}{17\cdot11\cdot3}=\dfrac{2}{3}\)
BN
1
NM
25 tháng 11 2021
Ta có:
=\(\dfrac{6x45x11x17}{15x12x34x33}\)
= \(\dfrac{6x15x3x11x17}{15x6x2x17x2x11x3}\)
= \(\dfrac{1}{2x2}\)
=\(\dfrac{1}{4}\)
Những chữ x ở trên là dấu nhân nhé
6 tháng 5 2015
\(\frac{7}{9}\cdot\frac{3}{5}=\frac{7\cdot3}{9\cdot5}=\frac{7\cdot1}{3\cdot5}=\frac{7}{15}\)
\(\frac{9}{22}\cdot55=\frac{9}{22}\cdot\frac{55}{1}=\frac{9\cdot55}{22\cdot1}=\frac{9\cdot5}{2\cdot1}=\frac{45}{2}\)
\(\frac{11}{17}:\frac{33}{34}=\frac{11}{17}\cdot\frac{34}{33}=\frac{11\cdot34}{17\cdot33}=\frac{1\cdot2}{1\cdot3}=\frac{2}{3}\)
DL
1
R
24 tháng 8 2019
Bài làm
\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]:\left[\frac{25}{4002}.\frac{2001}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{34}+\frac{33}{34}\right]:\left[\frac{1}{2}+\frac{9}{2}\right]\)
\(=\frac{1}{5}\)
VH
0
AT
0
TV
0
S
2
H9
HT.Phong (9A5)
CTVHS
30 tháng 6 2023
\(M=\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{2001}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(M=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{4002}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(M=\left(\dfrac{1}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right)\)
\(M=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)
\(M=1:5\)
\(M=\dfrac{1}{5}\)
30 tháng 6 2023
\(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)
=1/5
PT
0
11/17 : 33/34 = 11/17 x 34/33
= 11 x 17 x 2 / 17 x 11 x 3
= 2/3
mk có thể làm bài này theo cách rút gọn bạn nhé !
\(\frac{11}{17}:\frac{33}{34}=\frac{11}{17}\cdot\frac{34}{33}=\frac{2}{3}\)
. = x