Câu hỏi của Han Nguyễn

Question

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

Rayleigh · Accepted Answer

$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$Câu trả lời: $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$

Nguyễn Phúc Hưng · Answer

$\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$$\Leftrightarrow\left[{}\begin{matrix}x=1\x=-1\end{matrix}\right.$Câu trả lời: $\dfrac{2}{3}+\dfrac{1}{3}.x^2=\left(-2\right)^0$$\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.$

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