Whakaoti i te 6+(3-x)^2/x+2-1geq2-x^2/-x-2

$$\frac { 6 + ( 3 - x ) ^ { 2 } } { x + 2 } - 1 \geq \frac { 2 - x ^ { 2 } } { - x - 2 }$$

Whakaoti mō x

$$x\in (-2,\frac{15}{7}]$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(3-x\right)^{2}.

$$\frac{6+9-6x+x^{2}}{x+2}-1\geq \frac{2-x^{2}}{-x-2}$$

Tāpirihia te 6 ki te 9, ka 15.

$$\frac{15-6x+x^{2}}{x+2}-1\geq \frac{2-x^{2}}{-x-2}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 1 ki te \frac{x+2}{x+2}.

$$\frac{15-6x+x^{2}}{x+2}-\frac{x+2}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

Tā te mea he rite te tauraro o \frac{15-6x+x^{2}}{x+2} me \frac{x+2}{x+2}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{15-6x+x^{2}-\left(x+2\right)}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

Mahia ngā whakarea i roto o 15-6x+x^{2}-\left(x+2\right).

$$\frac{15-6x+x^{2}-x-2}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

Whakakotahitia ngā kupu rite i 15-6x+x^{2}-x-2.

$$\frac{13-7x+x^{2}}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

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

$$\frac{13-7x+x^{2}}{x+2}-\frac{2-x^{2}}{-x-2}\geq 0$$

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

$$\frac{13-7x+x^{2}}{x+2}-\frac{-\left(2-x^{2}\right)}{x+2}\geq 0$$

Tā te mea he rite te tauraro o \frac{13-7x+x^{2}}{x+2} me \frac{-\left(2-x^{2}\right)}{x+2}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{13-7x+x^{2}-\left(-\left(2-x^{2}\right)\right)}{x+2}\geq 0$$

Mahia ngā whakarea i roto o 13-7x+x^{2}-\left(-\left(2-x^{2}\right)\right).

$$\frac{13-7x+x^{2}+2-x^{2}}{x+2}\geq 0$$

Whakakotahitia ngā kupu rite i 13-7x+x^{2}+2-x^{2}.

$$\frac{15-7x}{x+2}\geq 0$$

Kia ≥0 rawa te otinga, hei ≤0 a 15-7x me x+2 tahi, hei ≥0 rānei rāua tahi, otiia tē taea ai hei kore te x+2. Whakaarohia te tauira ina tōraro tahi a 15-7x\leq 0 me x+2.

$$15-7x\leq 0 x+2<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina tōrunga tahi a 15-7x\geq 0 me x+2.

$$15-7x\geq 0 x+2>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-2,\frac{15}{7}\right].

$$x\in (-2,\frac{15}{7}]$$

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

$$x\in (-2,\frac{15}{7}]$$