Whakaoti i te (x+4)(x+3)-(2x-1)(x-5)

$$( x + 4 ) ( x + 3 ) - ( 2 x - 1 ) ( x - 5 )$$

Aromātai

$$7+18x-x^{2}$$

Ngā Upane Otinga

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o x+4 ki ia tau o x+3.

$$x^{2}+3x+4x+12-\left(2x-1\right)\left(x-5\right)$$

Pahekotia te 3x me 4x, ka 7x.

$$x^{2}+7x+12-\left(2x-1\right)\left(x-5\right)$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 2x-1 ki ia tau o x-5.

$$x^{2}+7x+12-\left(2x^{2}-10x-x+5\right)$$

Pahekotia te -10x me -x, ka -11x.

$$x^{2}+7x+12-\left(2x^{2}-11x+5\right)$$

Hei kimi i te tauaro o 2x^{2}-11x+5, kimihia te tauaro o ia taurangi.

$$x^{2}+7x+12-2x^{2}-\left(-11x\right)-5$$

Ko te tauaro o -11x ko 11x.

$$x^{2}+7x+12-2x^{2}+11x-5$$

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

$$-x^{2}+7x+12+11x-5$$

Pahekotia te 7x me 11x, ka 18x.

$$-x^{2}+18x+12-5$$

Tangohia te 5 i te 12, ka 7.

$$-x^{2}+18x+7$$

Whakaroha

$$7+18x-x^{2}$$

Ngā Upane Otinga

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o x+4 ki ia tau o x+3.

$$x^{2}+3x+4x+12-\left(2x-1\right)\left(x-5\right)$$

Pahekotia te 3x me 4x, ka 7x.

$$x^{2}+7x+12-\left(2x-1\right)\left(x-5\right)$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 2x-1 ki ia tau o x-5.

$$x^{2}+7x+12-\left(2x^{2}-10x-x+5\right)$$

Pahekotia te -10x me -x, ka -11x.

$$x^{2}+7x+12-\left(2x^{2}-11x+5\right)$$

Hei kimi i te tauaro o 2x^{2}-11x+5, kimihia te tauaro o ia taurangi.

$$x^{2}+7x+12-2x^{2}-\left(-11x\right)-5$$

Ko te tauaro o -11x ko 11x.

$$x^{2}+7x+12-2x^{2}+11x-5$$

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

$$-x^{2}+7x+12+11x-5$$

Pahekotia te 7x me 11x, ka 18x.

$$-x^{2}+18x+12-5$$

Tangohia te 5 i te 12, ka 7.

$$-x^{2}+18x+7$$