a) [tex]x^2+x+\frac{1}{4}=x+1-\sqrt{x+1}+\frac{1}{4}<=>(x+\frac{1}{2})^2=(\sqrt{x+1}-\frac{1}{2})^2[/tex]
<=>[TEX](x+\frac{1}{2})=(\sqrt{x+1}-\frac{1}{2})[/TEX] hoặc [TEX](x+\frac{1}{2})=-(\sqrt{x+1}-\frac{1}{2})[/TEX]
b)ĐKXĐ: x>0
Nhân quy đồng lên [tex]\sqrt{x}\sqrt{x+2}+x-2=0<=>x(x+2)=(x-2)^2[/tex] là pt bậc 2