Whakaoti i te (y+sqrt{6})(y-sqrt{6})

$$( y + \sqrt { 6 } ) ( y - \sqrt { 6 } )$$

Aromātai

$$y^{2}-6$$

Ngā Upane Otinga

Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$y^{2}-\left(\sqrt{6}\right)^{2}$$

Ko te pūrua o \sqrt{6} ko 6.

$$y^{2}-6$$

Kimi Pārōnaki e ai ki y

$$2y$$

Ngā Upane e Whakamahi ana i te Ture Pārōnaki mō te Tapeke

Whakaarohia te \left(y+\sqrt{6}\right)\left(y-\sqrt{6}\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y^{2}-\left(\sqrt{6}\right)^{2})$$

Ko te pūrua o \sqrt{6} ko 6.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y^{2}-6)$$

Ko te pārōnaki o tētahi pūrau ko te tapeke o ngā pārōnaki o ōna kīanga tau. Ko te pārōnaki o tētahi kīanga tau pūmau ko 0. Ko te pārōnaki o te ax^{n} ko te nax^{n-1}.

$$2y^{2-1}$$

Tango 1 mai i 2.

$$2y^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$2y$$