\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\)
=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}\)
=\(\frac{2}{4}\)
=\(\frac{1}{2}\)
Tinh A=\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\)
A=\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\)
\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}=?\)
\(\left(\frac{2}{3}+\frac{2}{5}-\frac{2}{9}\right):\left(\frac{4}{3}+\frac{4}{5}-\frac{4}{9}\right)\)
Thực hiện phep tính một các hợp lý:
\(\left(\frac{2}{3}+\frac{2}{5}-\frac{2}{9}\right):\left(\frac{4}{3}+\frac{4}{5}-\frac{4}{9}\right)-\left(3-\frac{3}{11}-\frac{3}{17}\right):\left(5-\frac{5}{11}-\frac{5}{17}\right)\)
Tính giá trị biểu thức
a,\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
b,\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
c,\(A=81\cdot\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right]:\frac{158158158}{711711711}\)
d,\(A=\frac{5\cdot\left(2^2.3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(\frac{3}{5}:\left(\frac{-1}{5}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\)
\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(-\frac{2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
Viết phân số đảo ngược của các phân số sau :\(\frac{2}{3};\frac{4}{7};\frac{3}{5};\frac{9}{4};\frac{10}{7};\frac{5}{8};4;\frac{1}{2};\frac{3}{8};5;\frac{1}{3};\frac{1}{9}.\)
\(S=\frac{1}{2}:\frac{3}{2}:\frac{4}{3}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}:\frac{9}{8}:\frac{10}{9}\)
