Whakaoti mō x
x\in (-\infty,-\frac{1}{5}]\cup [\frac{1}{5},\infty)
Graph
Tohaina
Kua tāruatia ki te papatopenga
-1+25x^{2}\geq 0
Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i 1-25x^{2}. I te mea he tōraro a -1, ka huri te ahunga koreōrite.
x^{2}\geq \frac{1}{25}
Me tāpiri te \frac{1}{25} ki ngā taha e rua.
x^{2}\geq \left(\frac{1}{5}\right)^{2}
Tātaitia te pūtakerua o \frac{1}{25} kia tae ki \frac{1}{5}. Tuhia anō te \frac{1}{25} hei \left(\frac{1}{5}\right)^{2}.
|x|\geq \frac{1}{5}
E mau ana te koreōrite mō |x|\geq \frac{1}{5}.
x\leq -\frac{1}{5}\text{; }x\geq \frac{1}{5}
Tuhia anō te |x|\geq \frac{1}{5} hei x\leq -\frac{1}{5}\text{; }x\geq \frac{1}{5}.
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