Question

a) \(\dfrac{ 2}{3 }\)+\(\dfrac{ 1}{3 }\).\(x^2\)=\((-2)^0\)

Answer 1

\(a,\dfrac{2}{3}+\dfrac{1}{3}\times x^2=\left(-2\right)^0\)

\(\dfrac{2}{3}+\dfrac{1}{3}\times x^2=1\)

\(\dfrac{1}{3}\times x^2=1-\dfrac{2}{3}\)

\(\dfrac{1}{3}\times x^2=\dfrac{1}{3}\)

\(x^2=\dfrac{1}{3}:\dfrac{1}{3}\)

\(x^2=1\)

\(\Rightarrow x=\pm1\)

Answer 2

\(\dfrac{2}{3}+\dfrac{1}{3}.x^2=\left(-2\right)^0\)

\(\Leftrightarrow\dfrac{1}{3}.x^2=1-\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{1}{3}x^2=\dfrac{1}{3}\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\left|x\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Answer 3

23+13×x2=123+13×�2=1

13×x2=1313×�2=13