Question

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

Answer 1

ĐKXĐ: \(-3\le x\le2\)

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

\(\Leftrightarrow x+3=3-x+2\sqrt{2-x}\)

\(\Leftrightarrow\sqrt{2-x}=x\left(x\ge0\right)\)

\(\Leftrightarrow2-x=x^2\left(x\ge0\right)\)

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

Answer 2

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

<=> \(\sqrt{x+3}=1+\sqrt{2-x}\)

<=> x + 3 = \(\left(1+\sqrt{2-x}\right)^2\)

<=> x + 3 = 1 + \(2\sqrt{2-x}+\left(2-x\right)\)

<=> x + 3 - 1 - 2 + x = \(2\sqrt{2-x}\)

<=> 2x = \(2\sqrt{2-x}\)

<=> 2x : 2 = \(\sqrt{2-x}\)

<=> x = \(\sqrt{2-x}\)

<=> x² = 2 - x

<=> x² + x = 2

<=> x(x + 1) = 2

Vì 2 là số nguyên tố nên chỉ có 1 tích duy nhất là 2 . 1

<=> \(\left[{}\begin{matrix}x=1\\x+1=2\\x=2\\x+1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\\x=2\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)