Whakaoti i te 5x(x-1)-3>2x-x

$$5 x ( x - 1 ) - 3 > 2 x - x$$

Whakaoti mō x

$$x\in \left(-\infty,\frac{3-2\sqrt{6}}{5}\right)\cup \left(\frac{2\sqrt{6}+3}{5},\infty\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te x-1.

$$5x^{2}-5x-3>2x-x$$

Pahekotia te 2x me -x, ka x.

$$5x^{2}-5x-3>x$$

Tangohia te x mai i ngā taha e rua.

$$5x^{2}-5x-3-x>0$$

Pahekotia te -5x me -x, ka -6x.

$$5x^{2}-6x-3>0$$

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.

$$5x^{2}-6x-3=0$$

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 5 mō te a, te -6 mō te b, me te -3 mō te c i te ture pūrua.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Mahia ngā tātaitai.

$$x=\frac{6±4\sqrt{6}}{10}$$

Whakaotia te whārite x=\frac{6±4\sqrt{6}}{10} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{2\sqrt{6}+3}{5} x=\frac{3-2\sqrt{6}}{5}$$

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

$$5\left(x-\frac{2\sqrt{6}+3}{5}\right)\left(x-\frac{3-2\sqrt{6}}{5}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{2\sqrt{6}+3}{5} me te x-\frac{3-2\sqrt{6}}{5}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{2\sqrt{6}+3}{5} me te x-\frac{3-2\sqrt{6}}{5}.

$$x-\frac{2\sqrt{6}+3}{5}<0 x-\frac{3-2\sqrt{6}}{5}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<\frac{3-2\sqrt{6}}{5}.

$$x<\frac{3-2\sqrt{6}}{5}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{2\sqrt{6}+3}{5} me te x-\frac{3-2\sqrt{6}}{5}.

$$x-\frac{3-2\sqrt{6}}{5}>0 x-\frac{2\sqrt{6}+3}{5}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{2\sqrt{6}+3}{5}.

$$x>\frac{2\sqrt{6}+3}{5}$$

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

$$x<\frac{3-2\sqrt{6}}{5}\text{; }x>\frac{2\sqrt{6}+3}{5}$$