Question

Tìm X ; a. \(\left(x-2\right)^2=1\) b. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\) c. \(\left(3.x+1\right)^3=-27\) d. \(\left(x+1\right)^{-2}=\dfrac{1}{9}\)

Answer 1

a, \(\left(x-2\right)^2=1\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=-1\\x-2=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

b, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

c, \(\left(3x+1\right)^3=-27\)

\(\Rightarrow3x+1=-3\Rightarrow3x=-4\Rightarrow x=-\dfrac{4}{3}\)

d, \(\left(x+1\right)^{-2}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=-3\\x+1=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

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