Bài 6 trang 13

Đề bài

Biết rằng ${10}^{\alpha }=2;{10}^{\beta }=5$.

Tính ${10}^{\alpha +\beta };{10}^{\alpha -\beta };{10}^{2\alpha };{10}^{-2\alpha };{1000}^{\beta };0,{01}^{2\alpha }$.

Lời giải chi tiết

$\begin{array}{c}{10}^{\alpha +\beta }={10}^{\alpha }{.10}^{\beta }=2.5=10\\ {10}^{\alpha -\beta }=\frac{{10}^{\alpha }}{{10}^{\beta }}=\frac{2}{5}\\ {10}^{2\alpha }={\left({10}^{\alpha }\right)}^2=2^2=4\\ {10}^{-2\alpha }=\frac{1}{{10}^{2\alpha }}=\frac{1}{4}\\ {1000}^{\beta }={\left({10}^3\right)}^{\beta }={\left({10}^{\beta }\right)}^3=5^3=125\\ 0,{01}^{2\alpha }={\left(\frac{1}{100}\right)}^{2\alpha }=\frac{1}{{100}^{2\alpha }}=\frac{1}{{\left({10}^2\right)}^{2\alpha }}=\frac{1}{{10}^{4\alpha }}=\frac{1}{{\left({10}^{\alpha }\right)}^4}=\frac{1}{2^4}=\frac{1}{16}\end{array}$