Question

giải phương trình \(\sqrt{x^2+x+\frac{1}{4}}=x\)

Answer 1

ĐK:\(x\ge0\)

\(\Leftrightarrow x^2+x+\dfrac{1}{4}=x^2\)

\(\Leftrightarrow x=\dfrac{-1}{4}\left(KTM\right)\)

Vậy pt vô nghiệm.

Answer 2

ĐK:x\(\ge0\)

\(\sqrt{x^2+x+\dfrac{1}{4}}=x\Leftrightarrow\sqrt{x^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}=x\Leftrightarrow\sqrt{\left(x+\dfrac{1}{2}\right)^2}=x\Leftrightarrow\left|x+\dfrac{1}{2}\right|=x\Leftrightarrow x+\dfrac{1}{2}=x\left(x\ge0\right)\Leftrightarrow\dfrac{1}{2}=0\left(ktm\right)\)

Vậy S=\(\varnothing\)