Whakaoti i te x^2/x-1leqx+1

$$\frac { x ^ { 2 } } { x - 1 } \leq x + 1$$

Whakaoti mō x

$$x<1$$

Ngā Upane Otinga

Tangohia te x mai i ngā taha e rua.

$$\frac{x^{2}}{x-1}-x\leq 1$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{x-1}{x-1}.

$$\frac{x^{2}}{x-1}-\frac{x\left(x-1\right)}{x-1}\leq 1$$

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

$$\frac{x^{2}-x\left(x-1\right)}{x-1}\leq 1$$

Mahia ngā whakarea i roto o x^{2}-x\left(x-1\right).

$$\frac{x^{2}-x^{2}+x}{x-1}\leq 1$$

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

$$\frac{x}{x-1}\leq 1$$

Kāore e taea a te tauraro x-1 te numa kore nā te mea kāore te rituatanga mā te kore e tautuhi. E rua ngā kēhi.

$$x-1>0 x-1<0$$

Whakaarohia te kēhi i ngā wā kei te tōrunga a x-1. Neke atu a -1 ki te taha matau.

$$x>1$$

Kāore te tōrite tuatahi e whakarerekē te aronga ina e whakarea ana mā x-1 mō x-1>0.

$$x\leq x-1$$

Neke atu ngā kīanga tau e whai ana i x ki te taha mauī me ētahi atu kupu katoa ki te taha matau.

$$x-x\leq -1$$

Pahekotia ngā kīanga tau ōrite.

$$0\leq -1$$

Whakaarohia te herenga x>1 e tautuhi ana ki runga.

$$x\in \emptyset$$

Tēnā, me whakaarohia te tauira ina e tōraro a x-1. Neke atu a -1 ki te taha matau.

$$x<1$$

E whakarerekē ana te aronga o te tōrite tuatahi hei ngā wā e whakarea ana a x-1 mō x-1<0.

$$x\geq x-1$$

Neke atu ngā kīanga tau e whai ana i x ki te taha mauī me ētahi atu kupu katoa ki te taha matau.

$$x-x\geq -1$$

Pahekotia ngā kīanga tau ōrite.

$$0\geq -1$$

Whakaarohia te herenga x<1 e tautuhi ana ki runga.

$$x<1$$

Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.

$$x<1$$