Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left[\left(0,5\right)^3\right]^x=\dfrac{1}{64}\\ \Leftrightarrow\left[\left(\dfrac{1}{2}\right)^3\right]^x=\dfrac{1}{64}\\ \Leftrightarrow\left(\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\\ Vậy:x=2\\ \rightarrow S=\left\{2\right\}\)
Ta có: \(\left[\left(-0.5\right)^3\right]^x=\dfrac{1}{64}\)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^{3x}=\dfrac{1}{64}\)
\(\Leftrightarrow3x=6\)
hay x=2
\(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
\(\Rightarrow\left(-\dfrac{1}{8}\right)^x=\left(-\dfrac{1}{8}\right)^2\)
\(\Rightarrow x=2\)
Ta có: \(\left[\left(-0.5\right)^3\right]^x=\dfrac{1}{64}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{3x}=\dfrac{1}{64}\)
\(\Leftrightarrow3x=6\)
hay x=2
[(-0.5)3]x = \(\frac{1}{64}\)
<=> \(\left(-\frac{1}{8}\right)^x\)= \(\left(-\frac{1}{8}\right)^2\)
<=> x = 2
\(\left[-\frac{1}{8}\right]^x=\frac{1}{64}\)
\(\frac{-1^x}{8^x}=\frac{1}{8^2}\)
\(\Rightarrow x=2\)
Vậy,........
a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
=\(-8+\frac{3}{2}-1-\frac{3}{10}\)
=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)
=-9+\(\frac{6}{5}\)
=\(\frac{-39}{5}\)
a.\(\frac{3}{5}+0,5-\frac{1}{4}+\frac{7}{2}\)/\(\frac{-6}{5}+\frac{1}{6}-\frac{13}{5}+0,5\)
=(\(\frac{3}{5}+\frac{1}{2}-\frac{1}{4}+\frac{7}{2}\)).60/(\(\frac{-6}{5}+\frac{1}{6}-\frac{13}{5}+\frac{1}{2}\)).60
=(36+30-15+210)/(-72+10-156+30)
=261/-188
Sửa lại nha, mk nhìn nhầm đề, -x = 1 thì x = -1 vậy là D nha
Chúc bn học tốt!
Câu hỏi là gì bn
Tìm x ạ