Question

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

Answer 1

\(Đk:x\ge0\)

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(\sqrt{x}-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\left(nhận\right)\)

- Vậy phương trình đã cho có nghiệm duy nhất \(x=1\)

Answer 2

⇔\(x^2-x+1+1-2\sqrt{x}=0\) 

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

⇔\(\left(x-1\right)^2\)\(-\left(x-1\right)\)=0

⇔\(\left(x-1\right)\)\(\left(x-2\right)=0\)

⇔x-1 = 0 hoặc x-2 = 0

+) x-1 = 0 ⇔ x=1

+) x-2 = 0 ⇔ x=2