3) b) [tex](x+2)^3-y^3+x-y+2=0<=>(x+2-y)\left [ (x+2)^2+y(x+2)+y^2+1 \right ]=0=>y=x+2[/tex]
[tex](x+2)^2+y(x+2)+y^2+1>0[/tex] => [tex](x+2)^2+y(x+2)+y^2+1=0[/tex] vô nghiệm
thay lên triển thôi
a) [TEX]5x(x+1)=3x\sqrt{2x^2+1}+12<=>5x(x+1)-12=3x\sqrt{2x^2+1}<=>(5x^2+5x-12)^2=9x^2(2x^2+1)[/TEX]
[TEX](x-2)(7x-6)(x^2+10x+12)=0[/TEX]=> x=..