Thực hiện phép tính:
\(\frac{0,5+\frac{2}{43}-\frac{2}{2017}}{\frac{3}{4}+\frac{3}{43}-\frac{3}{2017}}:\frac{\frac{4}{91}+\frac{4}{34}+\frac{4}{2017}}{\frac{5}{91}+\frac{5}{34}+\frac{5}{2017}}\)
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Những câu hỏi liên quan
XO
26 tháng 10 2019
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
XO
26 tháng 10 2019
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
3 tháng 3 2020
\(6.\left(-\frac{1}{3}\right)^2-\frac{5}{4}:0,5+3\frac{1}{2}\)
\(=6.\frac{1}{9}-\frac{5}{4}.2+\frac{7}{2}\)
\(=\frac{2}{3}-\frac{5}{2}+\frac{7}{2}\)
\(=-\frac{11}{6}+\frac{7}{2}\)
\(=\frac{5}{3}\)
\(\frac{2017}{2018}.\frac{15}{17}-\frac{32}{17}.\frac{2017}{2018}=\frac{2017}{2018}.\left(\frac{15}{17}-\frac{32}{17}\right)\)
\(=\frac{2017}{2108}.\left(-1\right)=-\frac{2017}{2018}\)
5 tháng 4 2017
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
19 tháng 6 2017
\(\frac{2017}{1.2.3}+\frac{2017}{2.3.4}+\frac{2017}{3.4.5}+...+\frac{2017}{19.20.21}\)
\(=2017\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\right)\)
TH
19 tháng 6 2017
\(=2017.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\right)\)
\(=2017.\left(1-\frac{1}{2}-\frac{1}{3}-\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)-...-\left(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\right)\right)\)
\(=2017.\left(1+\frac{1}{21}\right)\)phá ngoặc trước dấu trừ đổi dấu,rút gọn:
\(=2017.\frac{20}{21}=\frac{40340}{21}\)
23 tháng 4 2017
Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)
\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)
\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
Giờ ta thế vào bài toán ban đầu được
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)
\(=\frac{2017}{2018}\)
18 tháng 10 2018
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
18 tháng 10 2018
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
\(\frac{0,5+\frac{2}{43}-\frac{2}{2017}}{\frac{3}{4}+\frac{3}{43}-\frac{3}{2017}}:\frac{\frac{4}{91}+\frac{4}{34}+\frac{4}{2017}}{\frac{5}{91}+\frac{5}{34}+\frac{5}{2017}}\)
\(=\frac{\frac{2}{4}+\frac{2}{43}-\frac{2}{2017}}{\frac{3}{4}+\frac{3}{43}-\frac{3}{2017}}:\frac{4\left(\frac{1}{91}+\frac{1}{34}+\frac{1}{2017}\right)}{5\left(\frac{1}{91}+\frac{1}{34}+\frac{1}{2017}\right)}\)
\(=\frac{2\left(\frac{1}{4}+\frac{1}{43}-\frac{1}{2017}\right)}{3\left(\frac{1}{4}+\frac{1}{43}-\frac{1}{2017}\right)}:\frac{4}{5}\)
\(=\frac{2}{3}.\frac{5}{4}=\frac{5}{6}\)