c thiếu đề
d) (x+1/3)mũ 3 =1/27
\(\rightarrow\left(x+\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\rightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{1}{3}\\ \rightarrow x=0\)
c)(x+2) mũ 2 =36
\(\rightarrow\left(x+2\right)^2=6^2\\ \rightarrow x+2=6\\ x=6-2\\ \rightarrow x=4\)
\(d,\\ \left(x+\dfrac{1}{3}\right)^3=\dfrac{1}{27}\\ \Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^3\\ \Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\\ \Leftrightarrow x=0\\ ~\\ c,\\ \left(x+2\right)^2=36\\ \Leftrightarrow\left(x+2\right)^2=\left(\pm6\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
