Whakaoti i te k(4+sqrt{3})(4-sqrt{3})

$$k ( 4 + \sqrt { 3 } ) ( 4 - \sqrt { 3 } )$$

Aromātai

$$13k$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te k ki te 4+\sqrt{3}.

$$\left(4k+k\sqrt{3}\right)\left(4-\sqrt{3}\right)$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 4k+k\sqrt{3} ki ia tau o 4-\sqrt{3}.

$$16k-4\sqrt{3}k+4k\sqrt{3}-k\left(\sqrt{3}\right)^{2}$$

Pahekotia te -4\sqrt{3}k me 4k\sqrt{3}, ka 0.

$$16k-k\left(\sqrt{3}\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$16k-k\times 3$$

Whakareatia te -1 ki te 3, ka -3.

$$16k-3k$$

Pahekotia te 16k me -3k, ka 13k.

$$13k$$

Kimi Pārōnaki e ai ki k

$$13$$

Ngā Upane e Whakamahi ana i te Tautuhinga o te Pārōnaki

Whakamahia te āhuatanga tohatoha hei whakarea te k ki te 4+\sqrt{3}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\left(4k+k\sqrt{3}\right)\left(4-\sqrt{3}\right))$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 4k+k\sqrt{3} ki ia tau o 4-\sqrt{3}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(16k-4\sqrt{3}k+4k\sqrt{3}-k\left(\sqrt{3}\right)^{2})$$

Pahekotia te -4\sqrt{3}k me 4k\sqrt{3}, ka 0.

$$\frac{\mathrm{d}}{\mathrm{d}k}(16k-k\left(\sqrt{3}\right)^{2})$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{\mathrm{d}}{\mathrm{d}k}(16k-k\times 3)$$

Whakareatia te -1 ki te 3, ka -3.

$$\frac{\mathrm{d}}{\mathrm{d}k}(16k-3k)$$

Pahekotia te 16k me -3k, ka 13k.

$$\frac{\mathrm{d}}{\mathrm{d}k}(13k)$$

Ko te pārōnaki o ax^{n} ko nax^{n-1}.

$$13k^{1-1}$$

Tango 1 mai i 1.

$$13k^{0}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$13\times 1$$

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

$$13$$