Bài 2:
a) \(\frac{8^{14}}{4^{12}}\)
\(=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}\)
\(=\frac{2^{42}}{2^{24}}\)
\(=2^{18}\)
\(=262144.\)
b) \(\left(-\frac{1}{3}\right)^7.3^7\)
\(=\left[\left(-\frac{1}{3}\right).3\right]^7\)
\(=\left(-1\right)^7\)
\(=-1.\)
c) \(\frac{90^2}{15^2}\)
\(=\left(\frac{90}{15}\right)^2\)
\(=6^2\)
\(=36.\)
d) \(\frac{790^4}{79^4}\)
\(=\left(\frac{790}{79}\right)^4\)
\(=10^4\)
\(=10000.\)
Chúc bạn học tốt!
Mk làm tiếp cho bạn Vũ Minh Tuấn nhé!
Bài 1:
\(-\frac{64}{343}=x^3\)
\(\Rightarrow x^3=\left(-\frac{4}{7}\right)^3\)
\(\Rightarrow x=-\frac{4}{7}\)
Vậy \(x=-\frac{4}{7}\)
\(\left(x+20\right)^{100}+\left|y+4\right|=0\)
Ta có: \(\left(x+20\right)^{100}\ge0;\left|y+4\right|\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)
Vậy \(x=-20;y=-4\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}\)
