Question

giải phương trình: 4x²-2\(\sqrt{x+1}\)=x+2

Answer 1

ĐKXĐ: x>=-1

\(4x^2-2\sqrt{x+1}=x+2\)

=>\(4x^2-2\sqrt{x+1}-x-2=0\)

=>\(4x^2+3x-4x-3+1-2\sqrt{x+1}=0\)

=>\(\left(4x+3\right)\left(x-1\right)+1-\sqrt{4x+4}=0\)

=>\(\left(4x+3\right)\left(x-1\right)+\dfrac{1-4x-4}{1+\sqrt{4x+4}}=0\)

=>\(\left(4x+3\right)\left(x-1\right)-\dfrac{4x+3}{1+\sqrt{4x+4}}=0\)

=>\(\left(4x+3\right)\left(x-1-\dfrac{1}{1+\sqrt{4x+4}}\right)=0\)

=>4x+3=0

=>x=-3/4(nhận)