tìm x để AB có nghĩa A=căn bậc hai(x+2) * căn bậc hai(x-3) b=căn bậc hai(x+2)(x-3) 11/07/2021 Bởi Liliana tìm x để AB có nghĩa A=căn bậc hai(x+2) * căn bậc hai(x-3) b=căn bậc hai(x+2)(x-3)
$A = \sqrt {x + 2} .\sqrt {x – 3} $ có nghĩa thì : $\begin{array}{l} \left\{ \begin{array}{l} x + 2 \ge 0\\ x – 3 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge – 2\\ x \ge 3 \end{array} \right.\\ \Leftrightarrow x \ge 3 \end{array}$ $B = \sqrt {\left( {x + 2} \right)\left( {x – 3} \right)}$ có nghĩa thì: $\begin{array}{l} \left( {x + 2} \right)\left( {x – 3} \right) \ge 0\\ \Leftrightarrow \left[ \begin{array}{l} x \le – 2\\ x \ge 3 \end{array} \right. \end{array}$ $\Rightarrow \begin{array}{l} AB = \left( {\sqrt {x + 2} .\sqrt {x – 3} } \right).\sqrt {\left( {x + 2} \right)\left( {x – 3} \right)} \\ \left\{ \begin{array}{l} x \ge – 2\\ x \ge 3\\ \left[ \begin{array}{l} x \ge 3\\ x \le – 2 \end{array} \right. \end{array} \right. \Rightarrow x \ge 3 \end{array}$ $D = \left[ {3; + \infty } \right)$ Bình luận
Giải thích các bước giải: A = $\sqrt{x+2}$ . $\sqrt{x-3}$ Để A có nghĩa => $\left \{ {{x+2 ≥ 0} \atop {x-3 ≥ 0}} \right.$ => x ≥ 3 B = $\sqrt{(x+2)(x-3)}$ Để B có nghĩa => (x+2)(x-3) ≥0 \(\left[ \begin{array}{l}\left \{ {{x+2 ≥ 0} \atop {x-3 ≥0}} \right. \\\left \{ {{x+2 ≤0} \atop {x -3 ≤ 0}} \right. \end{array} \right.\) => \(\left[ \begin{array}{l}\left \{ {{x ≥ -2} \atop {x ≥ 3}} \right. \\\left \{ {{x ≤ -2} \atop {x ≤ 3}} \right. \end{array} \right.\) => \(\left[ \begin{array}{l}x ≥ 3\\x ≤ -2\end{array} \right.\) Bình luận
$A = \sqrt {x + 2} .\sqrt {x – 3} $ có nghĩa thì :
$\begin{array}{l} \left\{ \begin{array}{l} x + 2 \ge 0\\ x – 3 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge – 2\\ x \ge 3 \end{array} \right.\\ \Leftrightarrow x \ge 3 \end{array}$
$B = \sqrt {\left( {x + 2} \right)\left( {x – 3} \right)}$ có nghĩa thì:
$\begin{array}{l} \left( {x + 2} \right)\left( {x – 3} \right) \ge 0\\ \Leftrightarrow \left[ \begin{array}{l} x \le – 2\\ x \ge 3 \end{array} \right. \end{array}$
$\Rightarrow \begin{array}{l} AB = \left( {\sqrt {x + 2} .\sqrt {x – 3} } \right).\sqrt {\left( {x + 2} \right)\left( {x – 3} \right)} \\ \left\{ \begin{array}{l} x \ge – 2\\ x \ge 3\\ \left[ \begin{array}{l} x \ge 3\\ x \le – 2 \end{array} \right. \end{array} \right. \Rightarrow x \ge 3 \end{array}$
$D = \left[ {3; + \infty } \right)$
Giải thích các bước giải:
A = $\sqrt{x+2}$ . $\sqrt{x-3}$
Để A có nghĩa
=> $\left \{ {{x+2 ≥ 0} \atop {x-3 ≥ 0}} \right.$
=> x ≥ 3
B = $\sqrt{(x+2)(x-3)}$
Để B có nghĩa
=> (x+2)(x-3) ≥0
\(\left[ \begin{array}{l}\left \{ {{x+2 ≥ 0} \atop {x-3 ≥0}} \right. \\\left \{ {{x+2 ≤0} \atop {x -3 ≤ 0}} \right. \end{array} \right.\)
=> \(\left[ \begin{array}{l}\left \{ {{x ≥ -2} \atop {x ≥ 3}} \right. \\\left \{ {{x ≤ -2} \atop {x ≤ 3}} \right. \end{array} \right.\)
=> \(\left[ \begin{array}{l}x ≥ 3\\x ≤ -2\end{array} \right.\)