Whakaoti i te 1/sqrt{x}=1/2y+1/3z

$$\frac{ 1 }{ \sqrt{ x } } = \frac{ 1 }{ 2y } + \frac{ 1 }{ 3z }$$

Whakaoti mō y

$$y=-\frac{3\sqrt{x}z}{2\left(-3z+\sqrt{x}\right)}$$

$$z\neq 0\text{ and }\left(z<0\text{ or }x\neq 9z^{2}\right)\text{ and }x>0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe y ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 6yz, arā, te tauraro pātahi he tino iti rawa te kitea o 2y,3z.

$$6yzx^{-\frac{1}{2}}=3z+2y$$

Tangohia te 2y mai i ngā taha e rua.

$$6yzx^{-\frac{1}{2}}-2y=3z$$

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

$$\left(6zx^{-\frac{1}{2}}-2\right)y=3z$$

He hanga arowhānui tō te whārite.

$$\left(\frac{6z}{\sqrt{x}}-2\right)y=3z$$

Whakawehea ngā taha e rua ki te 6zx^{-\frac{1}{2}}-2.

$$\frac{\left(\frac{6z}{\sqrt{x}}-2\right)y}{\frac{6z}{\sqrt{x}}-2}=\frac{3z}{\frac{6z}{\sqrt{x}}-2}$$

Mā te whakawehe ki te 6zx^{-\frac{1}{2}}-2 ka wetekia te whakareanga ki te 6zx^{-\frac{1}{2}}-2.

$$y=\frac{3z}{\frac{6z}{\sqrt{x}}-2}$$

Whakawehe 3z ki te 6zx^{-\frac{1}{2}}-2.

$$y=\frac{3\sqrt{x}z}{2\left(3z-\sqrt{x}\right)}$$

Tē taea kia ōrite te tāupe y ki 0.

$$y=\frac{3\sqrt{x}z}{2\left(3z-\sqrt{x}\right)}\text{, }y\neq 0$$

Whakaoti mō x

$$x=36\times \left(\frac{yz}{2y+3z}\right)^{2}$$

$$\left(y>0\text{ and }y<-\frac{3z}{2}\right)\text{ or }\left(z>0\text{ and }y<-\frac{3z}{2}\right)\text{ or }\left(z>0\text{ and }y>0\right)$$