Whakaoti i te 15(x^2-1)>-16x

$$15 ( x ^ { 2 } - 1 ) > - 16 x$$

Whakaoti mō x

$$x\in \left(-\infty,-\frac{5}{3}\right)\cup \left(\frac{3}{5},\infty\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 15 ki te x^{2}-1.

$$15x^{2}-15>-16x$$

Me tāpiri te 16x ki ngā taha e rua.

$$15x^{2}-15+16x>0$$

Kia whakaotia te koreōrite, me tauwehe te taha mauī. Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$15x^{2}-15+16x=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 15 mō te a, te 16 mō te b, me te -15 mō te c i te ture pūrua.

$$x=\frac{-16±\sqrt{16^{2}-4\times 15\left(-15\right)}}{2\times 15}$$

Mahia ngā tātaitai.

$$x=\frac{-16±34}{30}$$

Whakaotia te whārite x=\frac{-16±34}{30} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{3}{5} x=-\frac{5}{3}$$

Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.

$$15\left(x-\frac{3}{5}\right)\left(x+\frac{5}{3}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{3}{5} me te x+\frac{5}{3}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{3}{5} me te x+\frac{5}{3}.

$$x-\frac{3}{5}<0 x+\frac{5}{3}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-\frac{5}{3}.

$$x<-\frac{5}{3}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{3}{5} me te x+\frac{5}{3}.

$$x+\frac{5}{3}>0 x-\frac{3}{5}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{3}{5}.

$$x>\frac{3}{5}$$

Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.

$$x<-\frac{5}{3}\text{; }x>\frac{3}{5}$$