Whakaoti i te 13/9x^2+1leqx^2+4/3x

$$\frac{ 13 }{ 9 } { x }^{ 2 } +1 \leq { x }^{ 2 } + \frac{ 4 }{ 3 } x$$

Whakaoti mō x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te x^{2} mai i ngā taha e rua.

$$\frac{13}{9}x^{2}+1-x^{2}\leq \frac{4}{3}x$$

Pahekotia te \frac{13}{9}x^{2} me -x^{2}, ka \frac{4}{9}x^{2}.

$$\frac{4}{9}x^{2}+1\leq \frac{4}{3}x$$

Tangohia te \frac{4}{3}x mai i ngā taha e rua.

$$\frac{4}{9}x^{2}+1-\frac{4}{3}x\leq 0$$

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.

$$\frac{4}{9}x^{2}+1-\frac{4}{3}x=0$$

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 \frac{4}{9} mō te a, te -\frac{4}{3} mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times \frac{4}{9}\times 1}}{\frac{4}{9}\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{\frac{4}{3}±0}{\frac{8}{9}}$$

He ōrite ngā whakatau.

$$x=\frac{3}{2}$$

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

$$\frac{4}{9}\left(x-\frac{3}{2}\right)^{2}\leq 0$$

E mau ana te koreōrite mō x=\frac{3}{2}.

$$x=\frac{3}{2}$$