Whakaoti i te x-1/x^2+3x+2+6/2+x-x^2-10-x/4-x^2

$$\frac { x - 1 } { x ^ { 2 } + 3 x + 2 } + \frac { 6 } { 2 + x - x ^ { 2 } } - \frac { 10 - x } { 4 - x ^ { 2 } }$$

Aromātai

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Ngā Hipanga Rongoā Poto

Tauwehea te x^{2}+3x+2. Tauwehea te 2+x-x^{2}.

$$\frac{x-1}{\left(x+1\right)\left(x+2\right)}+\frac{6}{\left(x-2\right)\left(-x-1\right)}-\frac{10-x}{4-x^{2}}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o \left(x+1\right)\left(x+2\right) me \left(x-2\right)\left(-x-1\right) ko \left(x-2\right)\left(x+1\right)\left(x+2\right). Whakareatia \frac{x-1}{\left(x+1\right)\left(x+2\right)} ki te \frac{x-2}{x-2}. Whakareatia \frac{6}{\left(x-2\right)\left(-x-1\right)} ki te \frac{-\left(x+2\right)}{-\left(x+2\right)}.

$$\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}+\frac{6\left(-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{10-x}{4-x^{2}}$$

Tā te mea he rite te tauraro o \frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} me \frac{6\left(-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{\left(x-1\right)\left(x-2\right)+6\left(-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{10-x}{4-x^{2}}$$

Mahia ngā whakarea i roto o \left(x-1\right)\left(x-2\right)+6\left(-1\right)\left(x+2\right).

$$\frac{x^{2}-2x-x+2-6x-12}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{10-x}{4-x^{2}}$$

Whakakotahitia ngā kupu rite i x^{2}-2x-x+2-6x-12.

$$\frac{x^{2}-9x-10}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{10-x}{4-x^{2}}$$

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{x^{2}-9x-10}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}.

$$\frac{\left(x-10\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{10-x}{4-x^{2}}$$

Me whakakore tahi te x+1 i te taurunga me te tauraro.

$$\frac{x-10}{\left(x-2\right)\left(x+2\right)}-\frac{10-x}{4-x^{2}}$$

Tauwehea te 4-x^{2}.

$$\frac{x-10}{\left(x-2\right)\left(x+2\right)}-\frac{10-x}{\left(x-2\right)\left(-x-2\right)}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o \left(x-2\right)\left(x+2\right) me \left(x-2\right)\left(-x-2\right) ko \left(x-2\right)\left(x+2\right). Whakareatia \frac{10-x}{\left(x-2\right)\left(-x-2\right)} ki te \frac{-1}{-1}.

$$\frac{x-10}{\left(x-2\right)\left(x+2\right)}-\frac{-\left(10-x\right)}{\left(x-2\right)\left(x+2\right)}$$

Tā te mea he rite te tauraro o \frac{x-10}{\left(x-2\right)\left(x+2\right)} me \frac{-\left(10-x\right)}{\left(x-2\right)\left(x+2\right)}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x-10-\left(-\left(10-x\right)\right)}{\left(x-2\right)\left(x+2\right)}$$

Mahia ngā whakarea i roto o x-10-\left(-\left(10-x\right)\right).

$$\frac{x-10+10-x}{\left(x-2\right)\left(x+2\right)}$$

Whakakotahitia ngā kupu rite i x-10+10-x.

$$\frac{0}{\left(x-2\right)\left(x+2\right)}$$

Ina whakawehea te kore ki te tau ehara i te kore ka kore tonu.

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Tauwehe

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