Question

[lê thị lan anh]

B= \(\left[\left(0,1\right)^2\right]^0\) +\(\left[\left(\frac{1}{7}\right)^{-1}\right]^2\) .\(\frac{1}{49}.\left[\left(2^2\right)^3.2^5\right]\)

Answer 1

[Nguyễn Huy Tú]

\(B=\left[\left(0,1\right)^2\right]^0+\left[\left(\frac{1}{7}\right)^{-1}\right]^2.\frac{1}{49}.\left[\left(2^2\right)^3.2^5\right]\)

\(=1+\left(\frac{1}{\frac{1}{7}}\right)^2.\frac{1}{49}.\left(2^6.2^5\right)\)

\(=1+7^2.\frac{1}{49}.2^{11}\)

\(\Rightarrow1+49.\frac{1}{49}.2^{11}\)

\(=1+2^{11}\)

Vậy \(B=1+2^{11}\)