Question

[Trần Anh Thư]

bai1 tìm x : \(\frac{-64}{343}=x^3\)

\(\left(x-20\right)^{100}+|y+4|=0\)

(\(x-\frac{1}{2})^3=\frac{1}{27}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

bai2 tinh: a) \(\frac{8^{14}}{4^{12}}\)

b)\(\left(\frac{-1}{3}\right)^7.3^7\)

c)\(\frac{90^2}{15^2}\)

d)\(\frac{790^4}{79^4}\)

Answer 1

[Vũ Minh Tuấn]

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!

Answer 2

[Can I Help You?]

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 \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}\)