Whakaoti i te 3x-A(A^3/A+A^2)=9-A^2

$$3 x - A ( \frac { A ^ { 3 } } { A + A ^ { 2 } } ) = 9 - A ^ { 2 }$$

Whakaoti mō x

$$x=-\frac{A^{2}-9A-9}{3\left(A+1\right)}$$

$$A\neq -1\text{ and }A\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te A\left(A+1\right).

$$3xA\left(A+1\right)-AA^{3}=A\left(A+1\right)\times 9-A^{2}A\left(A+1\right)$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 1 me te 3 kia riro ai te 4.

$$3xA\left(A+1\right)-A^{4}=A\left(A+1\right)\times 9-A^{2}A\left(A+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3xA ki te A+1.

$$3xA^{2}+3xA-A^{4}=A\left(A+1\right)\times 9-A^{2}A\left(A+1\right)$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 2 me te 1 kia riro ai te 3.

$$3xA^{2}+3xA-A^{4}=A\left(A+1\right)\times 9-A^{3}\left(A+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te A ki te A+1.

$$3xA^{2}+3xA-A^{4}=\left(A^{2}+A\right)\times 9-A^{3}\left(A+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te A^{2}+A ki te 9.

$$3xA^{2}+3xA-A^{4}=9A^{2}+9A-A^{3}\left(A+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -A^{3} ki te A+1.

$$3xA^{2}+3xA-A^{4}=9A^{2}+9A-A^{4}-A^{3}$$

Me tāpiri te A^{4} ki ngā taha e rua.

$$3xA^{2}+3xA=9A^{2}+9A-A^{4}-A^{3}+A^{4}$$

Pahekotia te -A^{4} me A^{4}, ka 0.

$$3xA^{2}+3xA=9A^{2}+9A-A^{3}$$

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

$$\left(3A^{2}+3A\right)x=9A^{2}+9A-A^{3}$$

He hanga arowhānui tō te whārite.

$$\left(3A^{2}+3A\right)x=9A+9A^{2}-A^{3}$$

Whakawehea ngā taha e rua ki te 3A^{2}+3A.

$$\frac{\left(3A^{2}+3A\right)x}{3A^{2}+3A}=\frac{A\left(9+9A-A^{2}\right)}{3A^{2}+3A}$$

Mā te whakawehe ki te 3A^{2}+3A ka wetekia te whakareanga ki te 3A^{2}+3A.

$$x=\frac{A\left(9+9A-A^{2}\right)}{3A^{2}+3A}$$

Whakawehe A\left(9A+9-A^{2}\right) ki te 3A^{2}+3A.

$$x=\frac{9+9A-A^{2}}{3\left(A+1\right)}$$

Whakaoti mō A (complex solution)

$$A=\frac{\sqrt{9x^{2}-66x+117}-3x+9}{2}$$

$$A=\frac{-\sqrt{9x^{2}-66x+117}-3x+9}{2}\text{, }x\neq 3$$

Whakaoti mō A

$$A=\frac{\sqrt{9x^{2}-66x+117}-3x+9}{2}$$

$$A=\frac{-\sqrt{9x^{2}-66x+117}-3x+9}{2}\text{, }x\geq \frac{13}{3}\text{ or }x<3$$