Whakaoti i te (x-1)(x^2+x+1)-9-2xleq(x-1)^3+x(3x-2)

$$( x - 1 ) ( x ^ { 2 } + x + 1 ) - 9 - 2 x \leq ( x - 1 ) ^ { 3 } + x ( 3 x - 2 )$$

Whakaoti mō x

$$x\geq -3$$

Ngā Upane Otinga

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te x^{2}+x+1 ka whakakotahi i ngā kupu rite.

$$x^{3}-1-9-2x\leq \left(x-1\right)^{3}+x\left(3x-2\right)$$

Tangohia te 9 i te -1, ka -10.

$$x^{3}-10-2x\leq \left(x-1\right)^{3}+x\left(3x-2\right)$$

Whakamahia te ture huarua \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} hei whakaroha \left(x-1\right)^{3}.

$$x^{3}-10-2x\leq x^{3}-3x^{2}+3x-1+x\left(3x-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te 3x-2.

$$x^{3}-10-2x\leq x^{3}-3x^{2}+3x-1+3x^{2}-2x$$

Pahekotia te -3x^{2} me 3x^{2}, ka 0.

$$x^{3}-10-2x\leq x^{3}+3x-1-2x$$

Pahekotia te 3x me -2x, ka x.

$$x^{3}-10-2x\leq x^{3}+x-1$$

Tangohia te x^{3} mai i ngā taha e rua.

$$x^{3}-10-2x-x^{3}\leq x-1$$

Pahekotia te x^{3} me -x^{3}, ka 0.

$$-10-2x\leq x-1$$

Tangohia te x mai i ngā taha e rua.

$$-10-2x-x\leq -1$$

Pahekotia te -2x me -x, ka -3x.

$$-10-3x\leq -1$$

Me tāpiri te 10 ki ngā taha e rua.

$$-3x\leq -1+10$$

Tāpirihia te -1 ki te 10, ka 9.

$$-3x\leq 9$$

Whakawehea ngā taha e rua ki te -3. I te mea he tōraro a -3, ka huri te ahunga koreōrite.

$$x\geq \frac{9}{-3}$$

Whakawehea te 9 ki te -3, kia riro ko -3.

$$x\geq -3$$