Whakaoti i te (2x+1)^2-4(x+1)^2leq-2

$$( 2 x + 1 ) ^ { 2 } - 4 ( x + 1 ) ^ { 2 } \leq - 2$$

Whakaoti mō x

$$x\geq -\frac{1}{4}$$

Ngā Upane Otinga

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

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te -4 ki te x^{2}+2x+1.

$$4x^{2}+4x+1-4x^{2}-8x-4\leq -2$$

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

$$4x+1-8x-4\leq -2$$

Pahekotia te 4x me -8x, ka -4x.

$$-4x+1-4\leq -2$$

Tangohia te 4 i te 1, ka -3.

$$-4x-3\leq -2$$

Me tāpiri te 3 ki ngā taha e rua.

$$-4x\leq -2+3$$

Tāpirihia te -2 ki te 3, ka 1.

$$-4x\leq 1$$

Whakawehea ngā taha e rua ki te -4. I te mea he tōraro a -4, ka huri te ahunga koreōrite.

$$x\geq -\frac{1}{4}$$