Whakaoti i te (x-3)^2geq(x-7)(x+5)

$$( x - 3 ) ^ { 2 } \geq ( x - 7 ) ( x + 5 )$$

Whakaoti mō x

$$x\leq 11$$

Ngā Upane Otinga

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

$$x^{2}-6x+9\geq \left(x-7\right)\left(x+5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-7 ki te x+5 ka whakakotahi i ngā kupu rite.

$$x^{2}-6x+9\geq x^{2}-2x-35$$

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

$$x^{2}-6x+9-x^{2}\geq -2x-35$$

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

$$-6x+9\geq -2x-35$$

Me tāpiri te 2x ki ngā taha e rua.

$$-6x+9+2x\geq -35$$

Pahekotia te -6x me 2x, ka -4x.

$$-4x+9\geq -35$$

Tangohia te 9 mai i ngā taha e rua.

$$-4x\geq -35-9$$

Tangohia te 9 i te -35, ka -44.

$$-4x\geq -44$$

Whakawehea ngā taha e rua ki te -4. I te mea he tōraro a -4, ka huri te ahunga koreōrite.

$$x\leq \frac{-44}{-4}$$

Whakawehea te -44 ki te -4, kia riro ko 11.

$$x\leq 11$$