Bài 52 trang 62 SBT toán 10 Cánh diều: a) (sqrt {8 - x}  + x =  - 4)...

Giải các phương trình sau:

a) $\sqrt {8 - x}  + x =  - 4$

b) $\sqrt {3{x^2} - 5x + 2}  + 3x = 4$

Bước 1: Đưa về PT dạng $\sqrt {f\left( x \right)}  = g\left( x \right)$

Bước 2:  $\sqrt {f\left( x \right)}  = g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) = {\left[ {g\left( x \right)} \right]^2}\end{array} \right.$

a) $\sqrt {8 - x}  + x =  - 4 \Leftrightarrow \sqrt {8 - x}  =  - x - 4$

$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} - x - 4 \ge 0\\8 - x = {\left( { - x - 4} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le  - 4\\8 - x = {x^2} + 8x + 16\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le  - 4\\{x^2} + 9x + 8 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le  - 4\\\left[ \begin{array}{l}x =  - 1\;(L)\\x =  - 8\;\end{array} \right.\end{array} \right.\quad  \Leftrightarrow x =  - 8\end{array}$

Vậy $S = \left\{ { - 8} \right\}$

b) $\sqrt {3{x^2} - 5x + 2}  + 3x = 4 \Leftrightarrow \sqrt {3{x^2} - 5x + 2}  = 4 - 3x$

$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}4 - 3x \ge 0\\3{x^2} - 5x + 2 = {\left( {4 - 3x} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{4}{3}\\3{x^2} - 5x + 2 = 9{x^2} - 24x + 16\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{4}{3}\\6{x^2} - 19x + 14 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{4}{3}\\\left[ \begin{array}{l}x = 2\;(L)\\x = \frac{7}{6}\;\end{array} \right.\end{array} \right.\quad  \Leftrightarrow x = \frac{7}{6}\;\end{array}$

Vậy $S = \left\{ {\frac{7}{6}} \right\}$