tính
\(1\frac{7}{5741}.\frac{1}{3759}-\frac{4}{3741}.1\frac{2}{5741}+\frac{1}{3759}+\frac{1}{3759.5741}\)
K
Khách
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Những câu hỏi liên quan
19 tháng 7 2022
Đặt 3759=a; 5741=b
Theo đề, ta có: \(E=4\dfrac{7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\left(1+\dfrac{2}{b}\right)+\dfrac{1}{a}+\dfrac{1}{ab}\)
\(=\dfrac{4b+7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\dfrac{b+2}{b}+\dfrac{b+1}{ab}\)
\(=\dfrac{4b+7-4b-8+b+1}{ab}=\dfrac{b}{ab}=\dfrac{1}{a}=\dfrac{1}{3759}\)
NT
20 tháng 5 2018
\(xy\left(x-y\right)+yz\left(y-z\right)+xz\left(z-x\right)\\ =xy\left(x-y\right)+yz\left[-\left(x-y\right)-\left(z-x\right)\right]+xz\left(z-x\right)\\ =xy\left(x-y\right)-yz\left(x-y\right)-yz\left(z-x\right)+xz\left(z-x\right)\\ =\left(x-y\right)\left(xy-yz\right)+\left(z-x\right)\left(xz-yz\right)\\ =y\left(x-y\right)\left(x-z\right)+z\left(z-x\right)\left(x-y\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
RM
0
H
8 tháng 3 2018
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right):\frac{919191}{808080}\)
\(=\left(\frac{1}{2}:4\right):\frac{919191}{808080}=\frac{1}{8}\cdot\frac{808080}{919191}=\frac{10}{91}\)
F
29 tháng 6 2019
Bài giải
\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)
\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)
MG
11 tháng 4 2018
=182.\(\orbr{\begin{cases}1.\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\\2.\left(\frac{1}{2}+\frac{1}{9}+\frac{1}{27}\right)\end{cases}}:\frac{4.\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1.\left(\frac{1}{3}+\frac{1}{49}-\frac{1}{343}\right)}:\frac{91}{80} \)
=.\(182.\left(\frac{1}{2}:\frac{4}{1}\right).\frac{91}{80}\)
=\(182.\frac{1}{8}.\frac{91}{80}\)
=.\(182.\frac{91}{640}\)
=\(\frac{8281}{320}\)
B
11 tháng 4 2018
\(=182.\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(=182.\frac{1}{8}.\frac{808080}{919191}=\frac{182}{8}.\frac{80}{91}=20\)
