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x^{2}\leq \frac{1}{5}
Whakawehea ngā taha e rua ki te 5. I te mea he tōrunga te 5, kāore e huri te ahunga koreōrite.
x^{2}\leq \left(\frac{\sqrt{5}}{5}\right)^{2}
Tātaitia te pūtakerua o \frac{1}{5} kia tae ki \frac{\sqrt{5}}{5}. Tuhia anō te \frac{1}{5} hei \left(\frac{\sqrt{5}}{5}\right)^{2}.
|x|\leq \frac{\sqrt{5}}{5}
E mau ana te koreōrite mō |x|\leq \frac{\sqrt{5}}{5}.
x\in \begin{bmatrix}-\frac{\sqrt{5}}{5},\frac{\sqrt{5}}{5}\end{bmatrix}
Tuhia anō te |x|\leq \frac{\sqrt{5}}{5} hei x\in \left[-\frac{\sqrt{5}}{5},\frac{\sqrt{5}}{5}\right].