Whakaoti i te +4y^2-8xy-2x+1=0

$$+ 4 y ^ { 2 } - 8 x y - 2 x + 1 = 0$$

Whakaoti mō x

$$x=\frac{4y^{2}+1}{2\left(4y+1\right)}$$

$$y\neq -\frac{1}{4}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te 4y^{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-8xy-2x+1=-4y^{2}$$

Tangohia te 1 mai i ngā taha e rua.

$$-8xy-2x=-4y^{2}-1$$

Pahekotia ngā kīanga tau katoa e whai ana i te x.

$$\left(-8y-2\right)x=-4y^{2}-1$$

Whakawehea ngā taha e rua ki te -8y-2.

$$\frac{\left(-8y-2\right)x}{-8y-2}=\frac{-4y^{2}-1}{-8y-2}$$

Mā te whakawehe ki te -8y-2 ka wetekia te whakareanga ki te -8y-2.

$$x=\frac{-4y^{2}-1}{-8y-2}$$

Whakawehe -4y^{2}-1 ki te -8y-2.

$$x=\frac{4y^{2}+1}{2\left(4y+1\right)}$$

Whakaoti mō y (complex solution)

$$y=\frac{\sqrt{4x^{2}+2x-1}}{2}+x$$

$$y=-\frac{\sqrt{4x^{2}+2x-1}}{2}+x$$

Whakaoti mō y

$$y=\frac{\sqrt{4x^{2}+2x-1}}{2}+x$$

$$y=-\frac{\sqrt{4x^{2}+2x-1}}{2}+x\text{, }x\geq \frac{\sqrt{5}-1}{4}\text{ or }x\leq \frac{-\sqrt{5}-1}{4}$$