Question

Giải các phương trình:

\(a,\sqrt{x-1}=x-1\)

\(b,x-\sqrt{2x+3}=0\)

Answer 1

a) \(x\in\left\{1;2\right\}\)

b) \(x=3\)

Answer 2

a, \(\sqrt{x-1}=x-1\) ( 1 )

ĐKXĐ: x - 1 ≥ 0 ⇒ x ≥ 1

( 1 ) ⇔ \(\sqrt{x-1}=x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-1=\left(x-1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x-1=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-3x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left(x-1\right)\left(x-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=1\left(N\right)\\x=2\left(N\right)\end{matrix}\right.\end{matrix}\right.\)

⇒ S = {1;2}

b, \(x-\sqrt{2x+3}=0\) ( 1 )

ĐKXĐ: 2x - 3 ≥ 0 ⇔ x ≥ \(\frac{3}{2}\)

( 1 ) ⇔ \(x=\sqrt{2x+3}\Leftrightarrow x^2=2x+3\)

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

⇔ ( x - 3)( x +1 ) = 0

⇔ \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(N\right)\\x=-1\left(L\right)\end{matrix}\right.\)

⇒ S = {3}