Whakaoti i te 1/x+2/frac{6{x}}

$$\frac { \frac { 1 } { x + 2 } } { \frac { 6 } { x } }$$

Aromātai

$$\frac{x}{6\left(x+2\right)}$$

Ngā Hipanga Rongoā Poto

Whakawehe \frac{1}{x+2} ki te \frac{6}{x} mā te whakarea \frac{1}{x+2} ki te tau huripoki o \frac{6}{x}.

$$\frac{x}{\left(x+2\right)\times 6}$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 6.

$$\frac{x}{6x+12}$$

Kimi Pārōnaki e ai ki x

$$\frac{1}{3\left(x+2\right)^{2}}$$

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

Whakawehe \frac{1}{x+2} ki te \frac{6}{x} mā te whakarea \frac{1}{x+2} ki te tau huripoki o \frac{6}{x}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x}{\left(x+2\right)\times 6})$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 6.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x}{6x+12})$$

Mō ngā pānga e rua e taea ana te pārōnaki, ko te pārōnaki o te otinga o ngā pānga e rua ko te tauraro whakareatia ki te pārōnaki o te taurunga tango i te taurunga whakareatia ki te pārōnaki o te tauraro, ā, ka whakawehea te katoa ki te tauraro kua pūruatia.

$$\frac{\left(6x^{1}+12\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{1})-x^{1}\frac{\mathrm{d}}{\mathrm{d}x}(6x^{1}+12)}{\left(6x^{1}+12\right)^{2}}$$

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}.

$$\frac{\left(6x^{1}+12\right)x^{1-1}-x^{1}\times 6x^{1-1}}{\left(6x^{1}+12\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{\left(6x^{1}+12\right)x^{0}-x^{1}\times 6x^{0}}{\left(6x^{1}+12\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{6x^{1}x^{0}+12x^{0}-x^{1}\times 6x^{0}}{\left(6x^{1}+12\right)^{2}}$$

Hei whakarea pū o te pūtake ōrite, tāpiri ana taupū.

$$\frac{6x^{1}+12x^{0}-6x^{1}}{\left(6x^{1}+12\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(6-6\right)x^{1}+12x^{0}}{\left(6x^{1}+12\right)^{2}}$$

Tango 6 mai i 6.

$$\frac{12x^{0}}{\left(6x^{1}+12\right)^{2}}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{12x^{0}}{\left(6x+12\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{12\times 1}{\left(6x+12\right)^{2}}$$

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

$$\frac{12}{\left(6x+12\right)^{2}}$$