Giải các phương trình sau:
a) \(\sqrt { – 4x + 4} = \sqrt { – {x^2} + 1} \)
b) \(\sqrt {3{x^2} – 6x + 1} = \sqrt {{x^2} – 3} \)
c) \(\sqrt {2x – 1} = 3x – 4\)
d) \(\sqrt { – 2{x^2} + x + 7} = x – 3\)
+ \(\sqrt {f\left( x \right)} = \sqrt {g\left( x \right)} \Leftrightarrow \left\{ \begin{array}{l}f\left( x \right) \ge 0\\f\left( x \right) = g\left( x \right)\end{array} \right.\)
+ \(\sqrt {f\left( x \right)} = g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) = {\left[ {g\left( x \right)} \right]^2}\end{array} \right.\)
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a) \(\sqrt { – 4x + 4} = \sqrt { – {x^2} + 1} \)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} – 4x + 4 \ge 0\\ – 4x + 4 = – {x^2} + 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 1\\{x^2} – 4x + 3 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 1\\\left[ \begin{array}{l}x = 1\\x = 3\;(L)\end{array} \right.\end{array} \right.\quad \Leftrightarrow x = 1\end{array}\)
Vậy \(S = \left\{ 1 \right\}\)
b) \(\sqrt {3{x^2} – 6x + 1} = \sqrt {{x^2} – 3} \)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}{x^2} – 3 \ge 0\\3{x^2} – 6x + 1 = {x^2} – 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} – 3 \ge 0\\2{x^2} – 6x + 4 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} – 3 \ge 0\\\left[ \begin{array}{l}x = 1\;(L)\\x = 2\end{array} \right.\end{array} \right.\quad \Leftrightarrow x = 2\end{array}\)
Vậy \(S = \left\{ 2 \right\}\)
c) \(\sqrt {2x – 1} = 3x – 4\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}3x – 4 \ge 0\\2x – 1 = {\left( {3x – 4} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{4}{3}\\2x – 1 = 9{x^2} – 24x + 16\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{4}{3}\\9{x^2} – 26x + 17 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{4}{3}\\\left[ \begin{array}{l}x = 1\;(L)\\x = \frac{{17}}{9}\end{array} \right.\end{array} \right. \Leftrightarrow x = \frac{{17}}{9}\end{array}\)
Vậy \(S = \left\{ {\frac{{17}}{9}} \right\}\)
d) \(\sqrt { – 2{x^2} + x + 7} = x – 3\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}x – 3 \ge 0\\ – 2{x^2} + x + 7 = {\left( {x – 3} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 3\\ – 2{x^2} + x + 7 = {x^2} – 6x + 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge 3\\3{x^2} – 7x + 2 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 3\\\left[ \begin{array}{l}x = 2\;(L)\\x = \frac{1}{3}\;(L)\end{array} \right.\end{array} \right.\end{array}\)
Vậy \(S = \emptyset \)