\(2^3+3.\frac{1}{9}-2^{-2}.\left[\left(-2\right)^2:\frac{1}{2}\right].8\)
\(=8+\frac{1}{3}-\frac{1}{4}.\left[4.4\right].8\)
\(=8+\frac{1}{12}.16.8\)
\(=8+\frac{4}{3}.8\)
\(=8.\left(\frac{4}{3}+1\right)\)
\(=8.\frac{7}{3}\)
\(=\frac{56}{3}\)
\(2^3+3.\frac{1}{9}-2^{-2}.\left[\left(-2\right)^2:\frac{1}{2}\right].8\)
\(=8+\frac{1}{3}-\frac{1}{4}.\left[4.4\right].8\)
\(=8+\frac{1}{12}.16.8\)
\(=8+\frac{4}{3}.8\)
\(=8.\left(\frac{4}{3}+1\right)\)
\(=8.\frac{7}{3}\)
\(=\frac{56}{3}\)
\(2^3+3\times\frac{1}{9}-2^{-2}\times\left(\left(-2\right)^2\div\frac{1}{2}\right)\times8\)
\(4\times\left(\frac{1}{4}\right)^2+25\times\left[\left(\frac{3}{4}\right)^3\div\left(\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)
\(2^3+3\times\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2\div\frac{1}{2}\right]-8\)
1. Rút gọn a. \(B=\frac{\left(-\frac{1}{3}\right)^3-\left(\frac{3}{4}\right)^3\times\left(-2\right)^2}{2\times\left(-1\right)^5+\left(\frac{3}{2}\right)^2-\frac{3}{8}}\) b.\(C=\frac{\left(\frac{1}{2}\right)^3\times2^{-2}\times8}{\left(-23\right)^2\times16}\)
Tính:
\(\left(-2\frac{1}{5}\right)\times\left(\frac{-9}{11}\right)\times\left(-1\frac{1}{14}\right)\times\frac{2}{5}+\left(\frac{-4}{7}+\frac{6}{11}\right)\div\)\(\frac{-3}{5}\)
\frac{\left(\frac{2}{5}\right)^7\times 5^{7\:}+\left(\frac{9}{4}\right)^3\div \left(\frac{3}{16}\right)^3}{2^7\times 5^2+512}
Gía trị của biểu thức: \(2^3+3\times\left(\frac{1}{2}\right)^0+8\times\left[\left(-2\right)^2\div\frac{1}{2}\right]\)
Viết các biểu thức sau dưới dạng \(a^n\)(a thuộc Q, n thuộc N)
a) \(9\times3^3\times\frac{1}{81}\times3^2\)
b) \(4\times2^5\div\left(2^3\times\frac{1}{16}\right)\)
c) \(3^2\times2^5\times\left(\frac{2}{3}\right)^2\)
d) \(\left(\frac{1}{3}\right)^2\times\frac{1}{9}\times9^2\)
Tính giá trị biểu thức: A=\(\frac{\left(1+17\right)\times\left(1+\frac{17}{2}\right)\times\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right)\times\left(1+\frac{19}{2}\right)\times\left(1+\frac{19}{3}\right)....\left(1+\frac{19}{17}\right)}\)
tính A biết
A=\(\left(\frac{1}{1+2}-1\right)\times\left(\frac{1}{1+2+3}-1\right)\times\left(\frac{1}{1+2+3+4}-1\right)\times...\times\left(\frac{1}{1+2+3+4+...+2018}-1\right)\)
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