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

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

Whakaoti mō x

$$x=\frac{1}{2}=0.5$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

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

Whakaarohia te \left(x-2\right)\left(x+2\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 2.

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

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

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

Tāpirihia te -4 ki te 3, ka -1.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x+1 ka whakakotahi i ngā kupu rite.

$$x^{2}-1+3x=3+x^{2}-x-2$$

Tangohia te 2 i te 3, ka 1.

$$x^{2}-1+3x=1+x^{2}-x$$

Tangohia te x^{2} mai i ngā taha e rua.

$$x^{2}-1+3x-x^{2}=1-x$$

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

$$-1+3x=1-x$$

Me tāpiri te x ki ngā taha e rua.

$$-1+3x+x=1$$

Pahekotia te 3x me x, ka 4x.

$$-1+4x=1$$

Me tāpiri te 1 ki ngā taha e rua.

$$4x=1+1$$

Tāpirihia te 1 ki te 1, ka 2.

$$4x=2$$

Whakawehea ngā taha e rua ki te 4.

$$x=\frac{2}{4}$$

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{2}$$