Question

\(\left(x+3\right)\sqrt{-x^2-8x+48}\)\(=x-24\)

Answer 1

Hình như đề bị sai hay sao ý. Tui nghĩ đề vậy nè:

Giải phương trình: \(\left(x+3\right)\sqrt{-x^2-x+48}=x-24\)

Đặt: \(u=\sqrt{-x^2-x+48}\) và \(v=x+3\left(u\ge0\right)\) ta suy ra:

\(\left\{{}\begin{matrix}u^2+v^2=-2x+57\\2ucv=2x-48\end{matrix}\right.\Rightarrow\left(u+v\right)^2=9\Rightarrow u+v=\pm3\)

+ Nếu \(u+v=3\) ta có:

\(\sqrt{-x^2-x+48}=-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\2x^2+8x-48=0\end{matrix}\right.\)

\(\Leftrightarrow x=-2-2\sqrt{7}\)

+ Nếu \(u+v=-3\) ta có:

\(\sqrt{-x^2-x+48}=-x-6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-6\\2x^2+8x-48=0\end{matrix}\right.\)

\(\Leftrightarrow x=-5-\sqrt{31}\)

Vậy phương trình có nghiệm: \(\left\{x=-2-2\sqrt{7};-5-\sqrt{31}\right\}\)

Answer 2

Điều kiện: \(\ - {x^2} - 8x + 48 \ge 0 \Leftrightarrow \left( {x - 4} \right)\left( {x + 12} \right) \le 0 \Leftrightarrow - 12 \le x \le 4.\)

\(\ PT \Leftrightarrow \left( {x + 3} \right)\sqrt { - {x^2} - 8x + 48} = - \dfrac{1}{2}{\left( {x + 3} \right)^2} - \dfrac{1}{2}\left( { - {x^2} - 8x + 48} \right) + \dfrac{9}{2}.\)

\(\ \Leftrightarrow {\left( {x + 3 + \sqrt { - {x^2} - 8x + 48} } \right)^2} = {3^2} \Leftrightarrow \left[ \begin{array}{l} \sqrt { - {x^2} - 8x + 48} = - x\\ \sqrt { - {x^2} - 8x + 48} = - x - 6 \end{array} \right.\)

- Nếu \(\ \sqrt { - {x^2} - 8x + 48} = - x \Leftrightarrow \left\{ \begin{array}{l} - 12 \le x \le 0\\ {x^2} + 4x - 24 = 0 \end{array} \right. \Leftrightarrow x = - 2\sqrt 7 - 2.\)

- Nếu \(\ \sqrt { - {x^2} - 8x + 48} = - x - 6 \Leftrightarrow \left\{ \begin{array}{l} - 12 \le x \le - 6\\ {x^2} + 10x - 6 = 0 \end{array} \right. \Leftrightarrow x = - \sqrt {31} - 5.\)

Vậy \(\ T = \left\{ { - \sqrt {31} - 5; - 2\sqrt 7 - 2} \right\}.\)

Answer 3

Điều kiện: $\ - {x^2} - 8x + 48 \ge 0 \Leftrightarrow \left( {x - 4} \right)\left( {x + 12} \right) \le 0 \Leftrightarrow - 12 \le x \le 4.$

$\ PT \Leftrightarrow \left( {x + 3} \right)\sqrt { - {x^2} - 8x + 48} = - \dfrac{1}{2}{\left( {x + 3} \right)^2} - \dfrac{1}{2}\left( { - {x^2} - 8x + 48} \right) + \dfrac{9}{2}.$
$\ \Leftrightarrow {\left( {x + 3 + \sqrt { - {x^2} - 8x + 48} } \right)^2} = {3^2} \Leftrightarrow \left[ \begin{array}{l}
\sqrt { - {x^2} - 8x + 48} = - x\\
\sqrt { - {x^2} - 8x + 48} = - x - 6

\end{array} \right.$
- Nếu $\ \sqrt { - {x^2} - 8x + 48} = - x \Leftrightarrow \left\{ \begin{array}{l}
- 12 \le x \le 0\\
{x^2} + 4x - 24 = 0
\end{array} \right. \Leftrightarrow x = - 2\sqrt 7 - 2.$
- Nếu $\ \sqrt { - {x^2} - 8x + 48} = - x - 6 \Leftrightarrow \left\{ \begin{array}{l}
- 12 \le x \le - 6\\
{x^2} + 10x - 6 = 0
\end{array} \right. \Leftrightarrow x = - \sqrt {31} - 5.$
Vậy $\ S = \left\{ { - \sqrt {31} - 5; - 2\sqrt 7 - 2} \right\}.$