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Question

Tính m, n, p biếta) $\left(\frac{1}{3}\right)^m=\frac{1}{81}$b) $\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5$c) $\left(-0,25\right)^p=\frac{1}{256}$

Võ Đông Anh Tuấn · Accepted Answer

a ) $\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4$$\Rightarrow m=4$b ) $\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5$       $\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5$       $\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5$       $\Leftrightarrow n=5$c ) $\left(-0,25\right)^p=\frac{1}{256}$   $\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}$   $\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4$   $\Leftrightarrow p=4$Câu trả lời: a ) $\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4$$\Rightarrow m=4$b ) $\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5$       $\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5$       $\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5$       $\Leftrightarrow n=5$c ) $\left(-0,25\right)^p=\frac{1}{256}$   $\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}$   $\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4$   $\Leftrightarrow p=4$

Trần Nguyễn Bảo Quyên · Answer

$a.$$\left(\frac{1}{3}\right)^m=\frac{1}{81}$$\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4$$\Rightarrow m=4$Vậy :        $m=4$$b.$$\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5$$\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}$$\Rightarrow n=5$Vậy :        $n=5$$c.$$\left(-0,25\right)^p=\frac{1}{256}$$\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}$$\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4$$\Rightarrow p=4$Vậy :        $p=4$ Câu trả lời:  $a.$$\left(\frac{1}{3}\right)^m=\frac{1}{81}$$\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4$$\Rightarrow m=4$Vậy :        $m=4$$b.$$\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5$$\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}$$\Rightarrow n=5$Vậy :        $n=5$$c.$$\left(-0,25\right)^p=\frac{1}{256}$$\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}$$\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4$$\Rightarrow p=4$Vậy :        $p=4$

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