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\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)

\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)

\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)

\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)

=> Làm nốt 

Mấy bài kia cũng làm tương tự

(- \(\dfrac{1}{3}\))³.\(x\) = \(\dfrac{1}{81}\)

          \(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))³

          \(x\) =  - (\(\dfrac{1}{3}\))⁴ :(\(\dfrac{1}{3}\))³

           \(x=-\dfrac{1}{3}\)

Vậy \(x=-\dfrac{1}{3}\)

a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)

\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)

\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)

\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)

\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)

b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)

\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)

\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)

\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)

\(\Leftrightarrow x=\frac{1}{-3}\)

c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

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

\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)

\(\Leftrightarrow x=\frac{5}{6}\)

d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)

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

\(\Leftrightarrow x=\frac{1}{6}\)

\(3^2.3^x.81-27^4=0\)

\(3^2.3^x.3^4-\left(3^3\right)^4=0\)

\(3^6.3^x-3^{12}=0\)

\(3^6.3^x-3^{12}=0\)

\(3^6.3^x=0+3^{12}\)

\(3^6.3^x=3^{12}\)

\(3^{6+x}=3^{12}\)

\(\Rightarrow6+x=12\)

\(x=12-6\)

\(\Rightarrow x=6\)

1: \(3^{x-1}=\frac{1}{243}\)

=>\(3^{x-1}=3^{-5}\)

=>x-1=-5

=>x=-4

2: \(81^{-2x}\cdot27^{x}=9^5\)

=>\(\left(3^4\right)^{-2x}\cdot\left(3^3\right)^{x}=3^{10}\)

=>\(3^{-8x+3x}=3^{10}\)

=>-5x=10

=>x=-2

3: \(\left(x-y+3\right)^2+\left(y-1\right)^2=0\)

=>x-y+3=0 và y-1=0

=>x=y-3 và y=1

=>y=1 và x=1-3=-2