Quay lại đáp án

$(x^3).y.(1 + y) + (x^2).(y^2).(2 + y) + x.(y^3) - 30 = 0$$⇔x^3.y + x^3.y^2 + 2.x^2.y^2 + x^2.y^3 + x.y^3 - 30 = 0$$⇔xy.(x^2 + y^2) + x ^2.y^2.(x + y) + 2.x^2.y^2 - 30 = 0$$⇔xy. [ (x+y)^2 - 2xy ] + x^2.y^2.(x + y) + 2.x^2.y^2 - 30 = 0$$⇔xy.(x+y)^2 + x^2.y^2.(x + y) - 30 = 0$$⇔xy.(x+y). (x+y+xy) = 30$$(x^2).y + x.(1 + y + y^2) + y - 11 = 0$$⇔x^2.y + x.y^2 + xy + x + y -11 = 0$$⇔xy.(x+y) + xy + (x+y) -11 = 0$Đặt $x+y=a;xy=b.$ Ta có:$\Rightarrow \left\{ \begin{array}{l} ab(a+b)=30\\ ab+(a+b)=11 \end{array} \right.$

Câu d$(x^3).y.(1 + y) + (x^2).(y^2).(2 + y) + x.(y^3) - 30 = 0$$⇔x^3.y + x^3.y^2 + 2.x^2.y^2 + x^2.y^3 + x.y^3 - 30 = 0$$⇔xy.(x^2 + y^2) + x ^2.y^2.(x + y) + 2.x^2.y^2 - 30 = 0$$⇔xy. [ (x+y)^2 - 2xy ] + x^2.y^2.(x + y) + 2.x^2.y^2 - 30 = 0$$⇔xy.(x+y)^2 + x^2.y^2.(x + y) - 30 = 0$$⇔xy.(x+y). (x+y+xy) = 30$$(x^2).y + x.(1 + y + y^2) + y - 11 = 0$$⇔x^2.y + x.y^2 + xy + x + y -11 = 0$$⇔xy.(x+y) + xy + (x+y) -11 = 0$Đặt $x+y=a;xy=b.$ Ta có:$\Rightarrow \left\{ \begin{array}{l} ab(a+b)=30\\ ab+(a+b)=11 \end{array} \right.$