Xet $x=0 y=0 khong la nghiem $ (1)$pt 1 <=> x^{3}-y^{6}+xy^{2}-y^{4}=0$$<=>(x-y^{2})(x^{2}+y^{4}+xy^{2}+y^{2})=0$$<=>x=y^{2} hoặc x^{2}+y^{4}+xy^{2}+y^{2}=0\Leftrightarrow (x+\frac{1}{2}y^{2})^{2}+\frac{3}{4}y^{4}+y^{2}= 0\Leftrightarrow x=y=0 $Thay $y^{2}=x$ vao pt2 ta co $\sqrt{4x+5}+\sqrt{x+8}=6$Cai nay bt roi nha. Minh lam den day thoi!!!!!!^^
Xet $x=0 y=0 khong la nghiem $ (1)$pt 1 <=> x^{3}-y^{6}+xy^{2}-y^{4}=0$$<=>(x-y^{2})(x^{2}+y^{4}+xy^{2}+y^{2})=0$$<=>x=y^{2} do x^{2}+y^{4}+xy^{2}+y^{2}=(x+\frac{1}{2}y^{2})^{2}+\frac{3}{4}y^{4}+y^{2}>0 vi (1)$Thay $y^{2}=x$ vao pt2 ta co $\sqrt{4x+5}+\sqrt{x+8}=6$Cai nay bt roi nha. Minh lam den day thoi!!!!!!^^
Xet $x=0 y=0 khong la nghiem $ (1)$pt 1 <=> x^{3}-y^{6}+xy^{2}-y^{4}=0$$<=>(x-y^{2})(x^{2}+y^{4}+xy^{2}+y^{2})=0$$<=>x=y^{2}
ho
ặc x^{2}+y^{4}+xy^{2}+y^{2}=
0\Leftrightarrow (x+\frac{1}{2}y^{2})^{2}+\frac{3}{4}y^{4}+y^{2}
= 0\Leftri
ghtarrow x=y=0 $Thay $y^{2}=x$ vao pt2 ta co $\sqrt{4x+5}+\sqrt{x+8}=6$Cai nay bt roi nha. Minh lam den day thoi!!!!!!^^