Question

Giải phương trình:

\(x^2+\sqrt{x^2-x+1}=x+1\)

Answer 1

\(x^2+\sqrt{x^2-x+1}=x+1\)(ĐKXĐ: \(x\in R\))

=>\(\sqrt{x^2-x+1}=x^2-x-1\)

=>\(x^2-x-1-\sqrt{x^2-x+1}=0\)

=>\(x^2-x+1-\sqrt{x^2-x+1}-2=0\)

=>\(\left(\sqrt{x^2-x+1}\right)^2-2\sqrt{x^2-x+1}+\sqrt{x^2-x+1}-2=0\)

=>\(\left(\sqrt{x^2-x+1}-2\right)\left(\sqrt{x^2-x+1}+1\right)=0\)

=>\(\sqrt{x^2-x+1}-2=0\)

=>\(\sqrt{x^2-x+1}=2\)

=>\(x^2-x+1=4\)

=>\(x^2-x-3=0\)

=>\(x=\dfrac{1\pm\sqrt{13}}{2}\)