Whakaoti i te (2sqrt{x}+sqrt{3})(2sqrt{x}-sqrt{3})

$$( 2 \sqrt { x } + \sqrt { 3 } ) ( 2 \sqrt { x } - \sqrt { 3 } )$$

Aromātai

$$4x-3$$

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}.

$$\left(2\sqrt{x}\right)^{2}-\left(\sqrt{3}\right)^{2}$$

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

$$2^{2}\left(\sqrt{x}\right)^{2}-\left(\sqrt{3}\right)^{2}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$4\left(\sqrt{x}\right)^{2}-\left(\sqrt{3}\right)^{2}$$

Tātaihia te \sqrt{x} mā te pū o 2, kia riro ko x.

$$4x-\left(\sqrt{3}\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$4x-3$$

Kimi Pārōnaki e ai ki x

$$4$$

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

Whakaarohia te \left(2\sqrt{x}+\sqrt{3}\right)\left(2\sqrt{x}-\sqrt{3}\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}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(2\sqrt{x}\right)^{2}-\left(\sqrt{3}\right)^{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(2^{2}\left(\sqrt{x}\right)^{2}-\left(\sqrt{3}\right)^{2})$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(4\left(\sqrt{x}\right)^{2}-\left(\sqrt{3}\right)^{2})$$

Tātaihia te \sqrt{x} mā te pū o 2, kia riro ko x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(4x-\left(\sqrt{3}\right)^{2})$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(4x-3)$$

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}.

$$4x^{1-1}$$

Tango 1 mai i 1.

$$4x^{0}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$4\times 1$$

Mō tētahi kupu t, t\times 1=t me 1t=t.

$$4$$