Bài 52 trang 62 SBT toán 10 – Cánh diều: Giải các phương trình sau: a) √8 – x + x = – 4 b) √3x^2 – 5x + 2 + 3x = 4

Đề bài/câu hỏi:

Giải các phương trình sau:

a) \(\sqrt {8 – x} + x = – 4\)

b) \(\sqrt {3{x^2} – 5x + 2} + 3x = 4\)

Hướng dẫn:

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

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

Lời giải:

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 \(S = \left\{ {\frac{7}{6}} \right\}\)