Whakaoti i te (7x+sqrt{22})*(7x-sqrt{22})=

$$( 7 x + \sqrt { 22 } ) \cdot ( 7 x - \sqrt { 22 } ) =$$

Aromātai

$$49x^{2}-22$$

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}.

$$\left(7x\right)^{2}-\left(\sqrt{22}\right)^{2}$$

Whakarohaina te \left(7x\right)^{2}.

$$7^{2}x^{2}-\left(\sqrt{22}\right)^{2}$$

Tātaihia te 7 mā te pū o 2, kia riro ko 49.

$$49x^{2}-\left(\sqrt{22}\right)^{2}$$

Ko te pūrua o \sqrt{22} ko 22.

$$49x^{2}-22$$

Kimi Pārōnaki e ai ki x

$$98x$$

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

Whakaarohia te \left(7x+\sqrt{22}\right)\left(7x-\sqrt{22}\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}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(7x\right)^{2}-\left(\sqrt{22}\right)^{2})$$

Whakarohaina te \left(7x\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(7^{2}x^{2}-\left(\sqrt{22}\right)^{2})$$

Tātaihia te 7 mā te pū o 2, kia riro ko 49.

$$\frac{\mathrm{d}}{\mathrm{d}x}(49x^{2}-\left(\sqrt{22}\right)^{2})$$

Ko te pūrua o \sqrt{22} ko 22.

$$\frac{\mathrm{d}}{\mathrm{d}x}(49x^{2}-22)$$

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}.

$$2\times 49x^{2-1}$$

Whakareatia 2 ki te 49.

$$98x^{2-1}$$

Tango 1 mai i 2.

$$98x^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$98x$$