Question

a) Tìm ĐKXĐ :

\(\sqrt{x^2-5x+6}\)

b) Giải phương trình :

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

Answer 1

a) ĐKXĐ: \(x^2-5x+6\ge0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\le2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)

b) ĐK: \(x\ge0\)

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

\(\Leftrightarrow x^2+4x+4-2=x^2\)

\(\Leftrightarrow4x+2=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Answer 2

Lê Thị Thục HiềnTrần Thanh Phương?Amanda?tthLightning Farron

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