Whakaoti i te x-1/x-2geq3/4-2x

$$\frac{ x-1 }{ x-2 } \geq \frac{ 3 }{ 4-2x }$$

Whakaoti mō x

$$x\in (-\infty,-\frac{1}{2}]\cup (2,\infty)$$

Ngā Upane Otinga

Tangohia te \frac{3}{4-2x} mai i ngā taha e rua.

$$\frac{x-1}{x-2}-\frac{3}{4-2x}\geq 0$$

Tauwehea te 4-2x.

$$\frac{x-1}{x-2}-\frac{3}{2\left(-x+2\right)}\geq 0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o x-2 me 2\left(-x+2\right) ko 2\left(x-2\right). Whakareatia \frac{x-1}{x-2} ki te \frac{2}{2}. Whakareatia \frac{3}{2\left(-x+2\right)} ki te \frac{-1}{-1}.

$$\frac{2\left(x-1\right)}{2\left(x-2\right)}-\frac{3\left(-1\right)}{2\left(x-2\right)}\geq 0$$

Tā te mea he rite te tauraro o \frac{2\left(x-1\right)}{2\left(x-2\right)} me \frac{3\left(-1\right)}{2\left(x-2\right)}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{2\left(x-1\right)-3\left(-1\right)}{2\left(x-2\right)}\geq 0$$

Mahia ngā whakarea i roto o 2\left(x-1\right)-3\left(-1\right).

$$\frac{2x-2+3}{2\left(x-2\right)}\geq 0$$

Whakakotahitia ngā kupu rite i 2x-2+3.

$$\frac{2x+1}{2\left(x-2\right)}\geq 0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x-2.

$$\frac{2x+1}{2x-4}\geq 0$$

Kia ≥0 rawa te otinga, hei ≤0 a 2x+1 me 2x-4 tahi, hei ≥0 rānei rāua tahi, otiia tē taea ai hei kore te 2x-4. Whakaarohia te tauira ina tōraro tahi a 2x+1\leq 0 me 2x-4.

$$2x+1\leq 0 2x-4<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\leq -\frac{1}{2}.

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

Whakaarohia te tauira ina tōrunga tahi a 2x+1\geq 0 me 2x-4.

$$2x+1\geq 0 2x-4>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>2.

$$x>2$$

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

$$x\leq -\frac{1}{2}\text{; }x>2$$