Whakaoti i te 3+4x/x^2+x-1=3/x-x/x+1

$$\frac { 3 + 4 x } { x ^ { 2 } + x } - 1 = \frac { 3 } { x } - \frac { x } { x + 1 }$$

Whakaoti mō x (complex solution)

$$x\in \mathrm{C}\setminus -1,0$$

Ngā Upane Otinga

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}+x,x,x+1.

$$3+4x+x\left(x+1\right)\left(-1\right)=\left(x+1\right)\times 3-xx$$

Whakareatia te x ki te x, ka x^{2}.

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}+x ki te -1.

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

Pahekotia te 4x me -x, ka 3x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 3.

$$3+3x-x^{2}=3x+3-x^{2}$$

Tangohia te 3x mai i ngā taha e rua.

$$3+3x-x^{2}-3x=3-x^{2}$$

Pahekotia te 3x me -3x, ka 0.

$$3-x^{2}=3-x^{2}$$

Tangohia te 3 mai i ngā taha e rua.

$$3-x^{2}-3=-x^{2}$$

Tangohia te 3 i te 3, ka 0.

$$-x^{2}=-x^{2}$$

Me tāpiri te x^{2} ki ngā taha e rua.

$$-x^{2}+x^{2}=0$$

Pahekotia te -x^{2} me x^{2}, ka 0.

$$0=0$$

Whakatauritea te 0 me te 0.

$$\text{true}$$

He pono tēnei mō tētahi x ahakoa.

$$x\in \mathrm{C}$$

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,0.

$$x\in \mathrm{C}\setminus -1,0$$

Whakaoti mō x

$$x\in \mathrm{R}\setminus -1,0$$

Ngā Upane Otinga

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}+x,x,x+1.

$$3+4x+x\left(x+1\right)\left(-1\right)=\left(x+1\right)\times 3-xx$$

Whakareatia te x ki te x, ka x^{2}.

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}+x ki te -1.

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

Pahekotia te 4x me -x, ka 3x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 3.

$$3+3x-x^{2}=3x+3-x^{2}$$

Tangohia te 3x mai i ngā taha e rua.

$$3+3x-x^{2}-3x=3-x^{2}$$

Pahekotia te 3x me -3x, ka 0.

$$3-x^{2}=3-x^{2}$$

Tangohia te 3 mai i ngā taha e rua.

$$3-x^{2}-3=-x^{2}$$

Tangohia te 3 i te 3, ka 0.

$$-x^{2}=-x^{2}$$

Me tāpiri te x^{2} ki ngā taha e rua.

$$-x^{2}+x^{2}=0$$

Pahekotia te -x^{2} me x^{2}, ka 0.

$$0=0$$

Whakatauritea te 0 me te 0.

$$\text{true}$$

He pono tēnei mō tētahi x ahakoa.

$$x\in \mathrm{R}$$

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,0.

$$x\in \mathrm{R}\setminus -1,0$$