Whakaoti i te {-5[9x^{6}y^{2}:(3x^{3}y)-3x^{3}y+9x^{2}y:(9xy)]^2+5x^2}^3+(4xy)^2:(-4xy)

$$\{ - 5 [ 9 x ^ { 6 } y ^ { 2 } : ( 3 x ^ { 3 } y ) - 3 x ^ { 3 } y + 9 x ^ { 2 } y : ( 9 x y ) ] ^ { 2 } + 5 x ^ { 2 } \} ^ { 3 } + ( 4 x y ) ^ { 2 } : ( - 4 x y )$$

Aromātai

$$-4xy$$

Ngā Upane Otinga

Me whakakore tahi te 3yx^{3} i te taurunga me te tauraro.

$$\left(-5\left(3yx^{3}-3x^{3}y+\frac{9x^{2}y}{9xy}\right)^{2}+5x^{2}\right)^{3}+\frac{\left(4xy\right)^{2}}{-4xy}$$

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

$$\left(-5\times \left(\frac{9x^{2}y}{9xy}\right)^{2}+5x^{2}\right)^{3}+\frac{\left(4xy\right)^{2}}{-4xy}$$

Me whakakore tahi te 9xy i te taurunga me te tauraro.

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

Pahekotia te -5x^{2} me 5x^{2}, ka 0.

$$0^{3}+\frac{\left(4xy\right)^{2}}{-4xy}$$

Tātaihia te 0 mā te pū o 3, kia riro ko 0.

$$0+\frac{\left(4xy\right)^{2}}{-4xy}$$

Whakarohaina te \left(4xy\right)^{2}.

$$0+\frac{4^{2}x^{2}y^{2}}{-4xy}$$

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$0+\frac{16x^{2}y^{2}}{-4xy}$$

Me whakakore tahi te 4xy i te taurunga me te tauraro.

$$0+\frac{4xy}{-1}$$

Ko te mea whakawehea ki te -1 ka hōmai i tōna kōaro.

$$0-4xy$$

Ko te tau i tāpiria he kore ka hua koia tonu.

$$-4xy$$

Whakaroha

$$-4xy$$

Ngā Upane Otinga

Me whakakore tahi te 3yx^{3} i te taurunga me te tauraro.

$$\left(-5\left(3yx^{3}-3x^{3}y+\frac{9x^{2}y}{9xy}\right)^{2}+5x^{2}\right)^{3}+\frac{\left(4xy\right)^{2}}{-4xy}$$

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

$$\left(-5\times \left(\frac{9x^{2}y}{9xy}\right)^{2}+5x^{2}\right)^{3}+\frac{\left(4xy\right)^{2}}{-4xy}$$

Me whakakore tahi te 9xy i te taurunga me te tauraro.

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

Pahekotia te -5x^{2} me 5x^{2}, ka 0.

$$0^{3}+\frac{\left(4xy\right)^{2}}{-4xy}$$

Tātaihia te 0 mā te pū o 3, kia riro ko 0.

$$0+\frac{\left(4xy\right)^{2}}{-4xy}$$

Whakarohaina te \left(4xy\right)^{2}.

$$0+\frac{4^{2}x^{2}y^{2}}{-4xy}$$

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$0+\frac{16x^{2}y^{2}}{-4xy}$$

Me whakakore tahi te 4xy i te taurunga me te tauraro.

$$0+\frac{4xy}{-1}$$

Ko te mea whakawehea ki te -1 ka hōmai i tōna kōaro.

$$0-4xy$$

Ko te tau i tāpiria he kore ka hua koia tonu.

$$-4xy$$