Whakaoti i te (x-2)^2>x(x+12)

$$( x - 2 ) ^ { 2 } > x ( x + 12 )$$

Whakaoti mō x

$$x<\frac{1}{4}$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-2\right)^{2}.

$$x^{2}-4x+4>x\left(x+12\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x+12.

$$x^{2}-4x+4>x^{2}+12x$$

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

$$x^{2}-4x+4-x^{2}>12x$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$-4x+4>12x$$

Tangohia te 12x mai i ngā taha e rua.

$$-4x+4-12x>0$$

Pahekotia te -4x me -12x, ka -16x.

$$-16x+4>0$$

Tangohia te 4 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-16x>-4$$

Whakawehea ngā taha e rua ki te -16. I te mea he tōraro a -16, ka huri te ahunga koreōrite.

$$x<\frac{-4}{-16}$$

Whakahekea te hautanga \frac{-4}{-16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te -4.

$$x<\frac{1}{4}$$