Whakaoti i te (x+5)(x+2)-3(x-4)^2>x(3-2x)+5

$$( x + 5 ) ( x + 2 ) - 3 ( x - 4 ) ^ { 2 } > x ( 3 - 2 x ) + 5$$

Whakaoti mō x

$$x>\frac{43}{28}$$

Ngā Upane Otinga

Whakamahia te āhuatanga tuaritanga hei whakarea te x+5 ki te x+2 ka whakakotahi i ngā kupu rite.

$$x^{2}+7x+10-3\left(x-4\right)^{2}>x\left(3-2x\right)+5$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-4\right)^{2}.

$$x^{2}+7x+10-3\left(x^{2}-8x+16\right)>x\left(3-2x\right)+5$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x^{2}-8x+16.

$$x^{2}+7x+10-3x^{2}+24x-48>x\left(3-2x\right)+5$$

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

$$-2x^{2}+7x+10+24x-48>x\left(3-2x\right)+5$$

Pahekotia te 7x me 24x, ka 31x.

$$-2x^{2}+31x+10-48>x\left(3-2x\right)+5$$

Tangohia te 48 i te 10, ka -38.

$$-2x^{2}+31x-38>x\left(3-2x\right)+5$$

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

$$-2x^{2}+31x-38>3x-2x^{2}+5$$

Tangohia te 3x mai i ngā taha e rua.

$$-2x^{2}+31x-38-3x>-2x^{2}+5$$

Pahekotia te 31x me -3x, ka 28x.

$$-2x^{2}+28x-38>-2x^{2}+5$$

Me tāpiri te 2x^{2} ki ngā taha e rua.

$$-2x^{2}+28x-38+2x^{2}>5$$

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

$$28x-38>5$$

Me tāpiri te 38 ki ngā taha e rua.

$$28x>5+38$$

Tāpirihia te 5 ki te 38, ka 43.

$$28x>43$$

Whakawehea ngā taha e rua ki te 28. I te mea he tōrunga te 28, kāore e huri te ahunga koreōrite.

$$x>\frac{43}{28}$$