\(\left(\frac{9}{11}-0,81\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}=0,00\left(81\right)^{2005}\)
\(\frac{1}{10^{4010}}=\frac{1}{100^{2005}}=\left(\frac{1}{100}\right)^{2005}=0,01^{2005}\)
Vì 0,00(81)<0,01 nên \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)
\(a.A=[\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}]+\frac{1890}{2005}+115\)
b.B=\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\cdot\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(42-5\frac{1}{3}\right)}\cdot\left(-1\frac{19}{93}\right)\right]\cdot\frac{31}{50}\)
Thực hiện phép tính: B= \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
thực hiện phép tính
\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)
Rút gọn
a)\(\frac{25^{5^{ }}.5^{10}}{100^5}.\frac{4^6.9^5+6^9.120}{8^4.3^{12^{ }}-6^{11}}\)
b)\(\left(\frac{4}{9}+\frac{1}{3}\right)^2+\left(\frac{3}{4}\right)^2:\left(\frac{3}{4}\right)^2:\left(-\frac{2}{3}\right)^3\)
c) (273 : 33) : \(\left[\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5\right]\)
CÁC BẠN GIÚP MIH VỚI
A=\(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)
B=\(\left(-6\right).10:\left[-0,25+\frac{1}{2}:\left(-2\right)\right]+1\frac{3}{4}\)
Thực hiện phép tính
\(\frac{5}{9}:\:\left(\frac{1}{11}-\frac{5}{22}\right)+\:\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
Tính:
A= \(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+3+...+2005+2006}\right)\)
Tính \(\text{A}\times\text{B}\), biết :
\(\text{A}=\frac{\frac{3}{10}+\frac{1}{2}-\frac{1}{6}}{\frac{1}{9}-\frac{1}{5}-\frac{1}{3}}\)
\(\text{B}=\frac{\left(3\cdot4\cdot2^{16}\right)^{^2}}{11\cdot2^{13}\cdot4^{11}-16^9}\)
B = \(\left(-\frac{5}{9}\right)\cdot\frac{3}{11}+\left(-\frac{13}{18}\right)\cdot\frac{3}{11}\)