## rut gon:a)(-48)^3:16^3 b)(-13/8)^3:(-32/13)^4 c)(9/10)^6:(17/-20)^6

Question

rut gon:a)(-48)^3:16^3
b)(-13/8)^3:(-32/13)^4
c)(9/10)^6:(17/-20)^6

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3 tháng 2021-09-07T15:12:11+00:00 2 Answers 41 views 0

1. Đáp án:

$$\begin{array}{l} a)\,\, – 27\\ b)\,\, – \frac{{{{13}^7}}}{{{2^{29}}}}\\ c)\,\,{\left( {\frac{{18}}{{17}}} \right)^6} \end{array}$$

Giải thích các bước giải:

$$\begin{array}{l} a)\,\,{\left( { – 48} \right)^3}:{16^3} = – {48^3}:{16^3} = – {\left( {\frac{{48}}{{16}}} \right)^3} = – {3^3} = – 27.\\ b)\,\,{\left( { – \frac{{13}}{8}} \right)^3}:{\left( { – \frac{{32}}{{13}}} \right)^4} = – \frac{{{{13}^3}}}{{{8^3}}}.\frac{{{{13}^4}}}{{{{32}^4}}} = – \frac{{{{13}^3}{{.13}^4}}}{{{{\left( {{2^3}} \right)}^3}.{{\left( {{2^5}} \right)}^4}}}\\ = – \frac{{{{13}^7}}}{{{2^9}{{.2}^{20}}}} = – \frac{{{{13}^7}}}{{{2^{29}}}}.\\ c)\,\,{\left( {\frac{9}{{10}}} \right)^6}:{\left( {\frac{{17}}{{ – 20}}} \right)^6} = \frac{{{9^6}}}{{{{10}^6}}}.\frac{{{{20}^6}}}{{{{17}^6}}}\\ = \frac{{{9^6}{{.2}^6}{{.10}^6}}}{{{{10}^6}{{.17}^6}}} = \frac{{{9^6}{{.2}^6}}}{{{{17}^6}}} = \frac{{{{18}^6}}}{{{{17}^6}}} = {\left( {\frac{{18}}{{17}}} \right)^6}. \end{array}$$

2. Đáp án:

Giải thích các bước giải:

a, 48=16.3

=> $\frac{(-48)^3}{16^3}$ = $\frac{-16^3.3^3}{16^3}$ =-3^3 = -27

b, Mình mạnh dạn sửa đề thành 2 phân số nhân với nhau mới rút ra số đẹp đc ^^

($\frac{-13}{8}$ )$^{3}$ . ($\frac{-32}{13}$ )$^{4}$

$\frac{-(13)^3}{8^3}$ . $\frac{(-32)^4}{13^4}$

= – $\frac{(8.4)^4}{13.8^3}$= – $\frac{8^4.4^4}{13.8^3}$

= – $\frac{8.4^4}{13}$=2048/13( cũng k đẹp lắm)

c, ($\frac{9}{10}$ )$^{6}$ : ($\frac{17}{-20}$ )$^{6}$

= $\frac{9^6}{10^6}$ : $\frac{17^6}{(-20)^6}$

= $\frac{9^6}{10^6}$ . $\frac{(-20)^6}{17^6}$

= $\frac{{9^6}2^6}{17^6}$

hay mình nhầm gì k b???