thực hiện phép tính:\(\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
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13 tháng 9 2016
Theo bài ra , ta có :
\(\frac{4}{5}\times\frac{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}\)
\(=\frac{4}{5}\times\frac{5\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{4\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(=\frac{4}{5}\times\frac{5}{4}\)
\(=1\)
Chúc bạn học tốt
NN
26 tháng 2 2021
\(\frac{4}{5}.\frac{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}=\frac{4}{5}.\frac{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}=\frac{4}{5}.\frac{5}{4}=1\)
12 tháng 5 2022
Sửa đề; \(\dfrac{1}{5}\cdot\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{1}{5}\cdot4:\dfrac{4}{5}=\dfrac{4}{5}\cdot\dfrac{5}{4}=1\)
2 tháng 7 2017
B=\(\frac{12+\frac{4}{3}-\frac{12}{37}-\frac{12}{53}}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12+\frac{12}{9}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{9}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}:\frac{4\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
=\(4:\frac{4}{5}\)
=\(5\)
AK
31 tháng 5 2018
P/s : Đề của bạn sai nên mik đã sửa lại rồi
Ta có :
\(B=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(\Rightarrow B=-\frac{6}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{1\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(\Rightarrow B=-\frac{6}{5}.4:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}.\frac{5}{4}\)
\(\Rightarrow B=-6\)
Vậy \(B=-6\)
~ Ủng hộ nhé
U
10 tháng 8 2018
\(B=-1\frac{1}{5}\cdot\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}\div\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=\frac{-6}{5}\cdot4\div\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(B=\frac{-24}{5}\div\frac{4}{5}\)
\(B=-6\)
10 tháng 8 2018
\(B=-1\frac{1}{5}.\frac{4.\frac{3}{7}}{\frac{3}{37}}:\frac{4+3.\frac{4}{1}}{5+3.\frac{5}{1}}\)
\(B=-\frac{6}{5}.\frac{148}{7}:\frac{4}{5}\)
\(B=-\frac{222}{7}\)
\(\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
= \(\frac{4.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
= \(\frac{4}{5}\)
\(\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}=\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}=\frac{4}{5}\)