Giải các phương trình sau:
a) \(\sqrt {7 - 2x} + x = 2\)
b) \(\sqrt { - 2{x^2} + 7x + 1} + 3x = 7\)
Bước 1: Đưa về PT dạng \(\sqrt {f\left( x \right)} = g\left( x \right)\)
Bước 2: \(\sqrt {f\left( x \right)} = g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) = {\left[ {g\left( x \right)} \right]^2}\end{array} \right.\)
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a) \(\sqrt {7 - 2x} + x = 2 \Leftrightarrow \sqrt {7 - 2x} = 2 - x\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}2 - x \ge 0\\7 - 2x = {\left( {2 - x} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le 2\\7 - 2x = {x^2} - 4x + 4\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 2\\{x^2} - 2x - 3 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le 2\\\left[ \begin{array}{l}x = 3\;(L)\\x = - 1\;\end{array} \right.\end{array} \right.\end{array}\)
Vậy \(S = \left\{ { - 1} \right\}\)
b) \(\sqrt { - 2{x^2} + 7x + 1} + 3x = 7 \Leftrightarrow \sqrt { - 2{x^2} + 7x + 1} = 7 - 3x\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}7 - 3x \ge 0\\ - 2{x^2} + 7x + 1 = {\left( {7 - 3x} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{7}{3}\\ - 2{x^2} + 7x + 1 = 9{x^2} - 42x + 49\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{7}{3}\\11{x^2} - 49x + 48 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{7}{3}\\\left[ \begin{array}{l}x = 3\;(L)\\x = \frac{{16}}{{11}}\;\end{array} \right.\quad \end{array} \right. \Leftrightarrow x = \frac{{16}}{{11}}\;\end{array}\)
Vậy \(S = \left\{ {\frac{{16}}{{11}}} \right\}\)