Quay lại đáp án

	2	thêm 214 ký tự trong đáp án		Sửa 29-03-16 08:38 PM ๖ۣۜTQT☾♋☽ 1
pt2 $:(xy-1)(x^2+y^2-2)=0$ TH1:thay $xy =1$ vào $pt 1.$ TH2:thay $x^2+y^2=2$ vào $pt1:$ $=2y^3-x^3+4x^2y-5xy^2\Leftrightarrow (y^3-x^3)+(y^3+4x^2y-5xy^2)\Leftrightarrow (2y-x)(y-x)^2=0$ .tự tìm ngiệm pt2 $:(xy-1)(x^2+y^2-2)=0$ TH1:thay $xy =1$ vào $pt 1.$ TH2:thay $x^2+y^2=2$ vào $pt1:$ pt2 $:(xy-1)(x^2+y^2-2)=0$ TH1:thay $xy =1$ vào $pt 1.$ TH2:thay $x^2+y^2=2$ vào $pt1:$ $=2y^3-x^3+4x^2y-5xy^2\Leftrightarrow (y^3-x^3)+(y^3+4x^2y-5xy^2)\Leftrightarrow (2y-x)(y-x)^2=0$ .tự tìm ngiệm

	1	tạo mới		Trả lời 29-03-16 08:21 PM ๖ۣۜTQT☾♋☽ 1
pt2 $:(xy-1)(x^2+y^2-2)=0$ TH1:thay $xy =1$ vào $pt 1.$ TH2:thay $x^2+y^2=2$ vào $pt1:$

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