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Câu hỏi: Giải các phương trình sau:
a) \(\sqrt {8 - x} + x = - 4\)
b) \(\sqrt {3{x^2} - 5x + 2} + 3x = 4\)
a) \(\sqrt {8 - x} + x = - 4\)
b) \(\sqrt {3{x^2} - 5x + 2} + 3x = 4\)
Phương pháp giải
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 chi tiết
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 \(S = \left\{ {\frac{7}{6}} \right\}\)
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 chi tiết
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 \(S = \left\{ {\frac{7}{6}} \right\}\)