Whakaoti i te (2x-1)^2geq(2x+3)^2

$$( 2 x - 1 ) ^ { 2 } \geq ( 2 x + 3 ) ^ { 2 }$$

Whakaoti mō x

$$x\leq -\frac{1}{2}$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2x-1\right)^{2}.

$$4x^{2}-4x+1\geq \left(2x+3\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+3\right)^{2}.

$$4x^{2}-4x+1\geq 4x^{2}+12x+9$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$4x^{2}-4x+1-4x^{2}\geq 12x+9$$

Pahekotia te 4x^{2} me -4x^{2}, ka 0.

$$-4x+1\geq 12x+9$$

Tangohia te 12x mai i ngā taha e rua.

$$-4x+1-12x\geq 9$$

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

$$-16x+1\geq 9$$

Tangohia te 1 mai i ngā taha e rua.

$$-16x\geq 9-1$$

Tangohia te 1 i te 9, ka 8.

$$-16x\geq 8$$

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

$$x\leq \frac{8}{-16}$$

Whakahekea te hautanga \frac{8}{-16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$x\leq -\frac{1}{2}$$