Question

Giải phương trình:

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

b. \(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)

Answer 1

Đk: \(2x^2-4x+1\ge0\)

pt đã cho \(\Leftrightarrow\left\{{}\begin{matrix}2x^2-4x+1=x^2-2x+1\\x\ge1\end{matrix}\right.\)

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

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

Kl: x=2

b) Đk: \(\left\{{}\begin{matrix}4x-20\ge0\\\dfrac{x-5}{9}\ge0\\1-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x\le1\end{matrix}\right.\) (vô lý)

Vậy ptvn

P/s: câu b hơi kỳ