Câu hỏi của terminat0r4

Question

(x + 1/2)^2=1/6tính

Võ Đông Anh Tuấn · Accepted Answer

$\left(x+\frac{1}{2}\right)^2=\frac{1}{6}$$\left(x+\frac{1}{2}\right)=+-\sqrt{\frac{1}{6}}$Khi : $\left(x+\frac{1}{2}\right)=\frac{1}{6}$$x=\frac{1}{6}-\frac{1}{2}$$x=-\frac{1}{3}$Khi$\left(x+\frac{1}{2}\right)=-\frac{1}{6}$$x=-\frac{1}{6}-\frac{1}{2}$$x=-\frac{2}{3}$Câu trả lời: $\left(x+\frac{1}{2}\right)^2=\frac{1}{6}$$\left(x+\frac{1}{2}\right)=+-\sqrt{\frac{1}{6}}$Khi : $\left(x+\frac{1}{2}\right)=\frac{1}{6}$$x=\frac{1}{6}-\frac{1}{2}$$x=-\frac{1}{3}$Khi$\left(x+\frac{1}{2}\right)=-\frac{1}{6}$$x=-\frac{1}{6}-\frac{1}{2}$$x=-\frac{2}{3}$

Dương Lam Hàng · Answer

$\left(x+\frac{1}{2}\right)^2=\frac{1}{6}$$\Rightarrow x^2+\left(\frac{1}{2}\right)^2=\frac{1}{6}$$\Rightarrow x^2+\frac{1}{4}=\frac{1}{6}$$\Rightarrow x^2=\frac{1}{6}-\frac{1}{4}=\frac{-1}{12}$$\Rightarrow x^2=\frac{-1}{12}$Câu trả lời: $\left(x+\frac{1}{2}\right)^2=\frac{1}{6}$$\Rightarrow x^2+\left(\frac{1}{2}\right)^2=\frac{1}{6}$$\Rightarrow x^2+\frac{1}{4}=\frac{1}{6}$$\Rightarrow x^2=\frac{1}{6}-\frac{1}{4}=\frac{-1}{12}$$\Rightarrow x^2=\frac{-1}{12}$

Nguyễn Tuấn Tài · Answer

$\left(x+\frac{1}{2}\right)^2=\frac{1}{6}$=>$x^2+\frac{1}{2}^2=\frac{1}{6}$$x^2+\frac{1}{4}=\frac{1}{6}$$x^2=\frac{1}{6}-\frac{1}{4}=-\frac{1}{12}$$x=\sqrt{-\frac{1}{12}}$Câu trả lời: $\left(x+\frac{1}{2}\right)^2=\frac{1}{6}$=>$x^2+\frac{1}{2}^2=\frac{1}{6}$$x^2+\frac{1}{4}=\frac{1}{6}$$x^2=\frac{1}{6}-\frac{1}{4}=-\frac{1}{12}$$x=\sqrt{-\frac{1}{12}}$

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