Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{72^2}{24^2}=\frac{3^2.24^2}{24^2}=3^2=9\)
\(\frac{\left(-7,5\right)^2}{2,5^2}=\frac{\left(7,5\right)^2}{\left(2,5\right)^2}=\frac{\left(2,5\right)^2.3^2}{\left(2,5\right)^2}=9\)
\(\frac{15^2}{27}=\frac{3^2.5^2}{27}=\frac{9.25}{27}=\frac{25}{3}\)
\(\frac{72^2}{24^2}=\left(\frac{72}{24}\right)^2=3^2=9\)
\(\frac{\left(-7,5\right)^3}{\left(2,5\right)^3}=\left(\frac{-7,5}{2,5}\right)^3=\left(-3\right)^3=-27\)
\(\frac{15^3}{27}=\frac{15^3}{3^3}=\left(\frac{15}{3}\right)^3=5^3=125\)
Chúc bạn hok tốt 
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
\(A=\left|4x-3\right|+\left|5y+7,5\right|+10\)
Mà \(\left|4x-3\right|\ge0\)với mọi x
\(\left|5y+7,5\right|\ge0\)với mọi y
\(\Rightarrow A\)có GTNN là 10
Để A có GTNN thì :
\(4x-3=0\) \(5y+7,5=0\)
\(4x=3\) \(5y=-7,5\)
\(x=\frac{3}{4}\) \(y=-1,5\)
\(B=\frac{5,8}{\left|2,5-x\right|+5,8}\)
Mà \(\left|2,5-x\right|\ge0\)
\(\Rightarrow\)GTNN \(\left|2,5-x\right|+5,8=5,8\)
Để B có GTLN \(\Rightarrow2,5-x=0\)
\(\Rightarrow x=2,5\)
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 \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}\)
\(\frac{72^2}{24^2}=\frac{72}{24}=3\)
\(\frac{\left(-7,5\right)^3}{\left(2,5\right)^3}=\frac{-7,5}{2,5}=-3\)
\(\frac{15^3}{27}=\frac{\left(3.5\right)^3}{\left(3.9\right)}=\frac{3^3.5^3}{3.9}=\frac{3^2.5^3}{9}=\frac{9.5^3}{9}=5^3=125\)