Question

giải phương trình \(\sqrt{2x-1}+2=x\)

Answer 1

\(x=5\)

Answer 2

\(\Leftrightarrow\sqrt{2x-1}=x-2\left(x\ge2\right)\\ \Leftrightarrow2x-1=x^2-4x+4\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\left(tm\right)\)

Answer 3

\(\sqrt{2x-1}+2=x\left(ĐK:x\ge\dfrac{1}{2}\right)\)

<=> \(\sqrt{2x-1}=x-2\)

<=> \(\left\{{}\begin{matrix}x\ge2\\2x-1=x^2-4x+4\left(1\right)\end{matrix}\right.\)

(1) <=> \(x^2-6x+5=0< =>\left(x-1\right)\left(x-5\right)=0\)

<=> \(\left\{{}\begin{matrix}x=1\left(L\right)\\x=5\left(C\right)\end{matrix}\right.\)