Question

1. Tìm TXĐ, TGT của hàm số: y=\(\sqrt{x+2}+\sqrt{2-x}\)

Answer 1

`@` H/s xác định `<=>{(x+2 >= 0),(2-x >= 0):}<=>{(x >= -2),(x <= 2):}<=>-2 <= x <= 2`

   `=>TXĐ: D=[-2;2]`

`@-2 <= x <= 2`

`<=>{(0 <= x+2 <= 4),(2 >= -x >= -2):}`

`<=>{(0 <= x+2 <= 4),(4 >= 2-x >= 0):}`

`<=>{(0 <= \sqrt{x+2} <= 2),(2 >= \sqrt{2-x} >= 0):}`

   `=>TGT` là `[0;2]`

Answer 2

\(y=\sqrt{x+2}+\sqrt{2-x}\)

y có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x+2>0\\2-x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x>2\end{matrix}\right.\)

TXD D = \(\left(2;+\infty\right)\)

Answer 3

\(đk\left\{{}\begin{matrix}x+2\ge0\\2-x\ge0\end{matrix}\right.=>\left\{{}\begin{matrix}x\ge-2\\x\le2\end{matrix}\right.\)

\(=>TXĐ:\left[-2;2\right]\)