Whakaoti i te 1/x=1/u+1/v

$$\frac{ 1 }{ x } = \frac{ 1 }{ u } + \frac{ 1 }{ v }$$

Whakaoti mō u

$$u=-\frac{vx}{x-v}$$

$$v\neq 0\text{ and }x\neq 0\text{ and }x\neq v$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe u ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te uvx, arā, te tauraro pātahi he tino iti rawa te kitea o x,u,v.

$$uv=vx+ux$$

Tangohia te ux mai i ngā taha e rua.

$$uv-ux=vx$$

Pahekotia ngā kīanga tau katoa e whai ana i te u.

$$\left(v-x\right)u=vx$$

Whakawehea ngā taha e rua ki te -x+v.

$$\frac{\left(v-x\right)u}{v-x}=\frac{vx}{v-x}$$

Mā te whakawehe ki te -x+v ka wetekia te whakareanga ki te -x+v.

$$u=\frac{vx}{v-x}$$

Tē taea kia ōrite te tāupe u ki 0.

$$u=\frac{vx}{v-x}\text{, }u\neq 0$$

Whakaoti mō v

$$v=-\frac{ux}{x-u}$$

$$u\neq 0\text{ and }x\neq 0\text{ and }x\neq u$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe v ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te uvx, arā, te tauraro pātahi he tino iti rawa te kitea o x,u,v.

$$uv=vx+ux$$

Tangohia te vx mai i ngā taha e rua.

$$uv-vx=ux$$

Pahekotia ngā kīanga tau katoa e whai ana i te v.

$$\left(u-x\right)v=ux$$

Whakawehea ngā taha e rua ki te -x+u.

$$\frac{\left(u-x\right)v}{u-x}=\frac{ux}{u-x}$$

Mā te whakawehe ki te -x+u ka wetekia te whakareanga ki te -x+u.

$$v=\frac{ux}{u-x}$$

Tē taea kia ōrite te tāupe v ki 0.

$$v=\frac{ux}{u-x}\text{, }v\neq 0$$