Advertisements (Quảng cáo)
Giải các phương trình :
a) \(\sqrt {11x – 8} = 6\);
b) \(\sqrt {2x + 1} + 1 = x\);
c) \(2\sqrt {x – 1} + \dfrac{1}{3}\sqrt {9x – 9} = 15\);
d) \(3\sqrt {27x} – 2\sqrt {12x} – 5 = 10\);
e) \(\sqrt {{x^2} – 12x + 36} + 3 = 10\);
f) \(\sqrt {x + 3 + 4\sqrt {x – 1} } + \sqrt {x + 8 – 6\sqrt {x – 1} } = 5\).
+) Tìm ĐKXĐ của x.
+) Sử dụng các công thức biến đổi căn bậc hai để giải phương trình.
|
\(a)\;\sqrt {11x – 8} = 6\;\) Điều kiện: \(11x – 8 \ge 0 \Leftrightarrow x \ge \dfrac{8}{{11}}.\) \(\begin{array}{l}Pt \Leftrightarrow 11x – 8 = 36\\ \Leftrightarrow 11x = 44\\ \Leftrightarrow x = 4\;\;\left( {tm} \right).\end{array}\) Vậy \(x = 4.\) \(\begin{array}{l}c)\;2\sqrt {x – 1} + \dfrac{1}{3}\sqrt {9x – 9} = 15\;\\ \Leftrightarrow 2\sqrt {x – 1} + \dfrac{1}{3}\sqrt {9\left( {x – 1} \right)} = 15\end{array}\) Điều kiện: \(x – 1 \ge 0 \Leftrightarrow x \ge 1.\) \(\begin{array}{l}Pt \Leftrightarrow 2\sqrt {x – 1} + \sqrt {x – 1} = 15\\ \Leftrightarrow 3\sqrt {x – 1} = 15\\ \Leftrightarrow \sqrt {x – 1} = 5\\ \Leftrightarrow x – 1 = 25\\ \Leftrightarrow x = 26.\end{array}\) Advertisements (Quảng cáo) Vậy \(x = 26.\) \(e)\;\;\sqrt {{x^2} – 12x + 36} + 3 = 10\) Điều kiện: \({x^2} – 12x + 36 \ge 0 \Leftrightarrow {\left( {x – 6} \right)^2} \ge 0\;\;\forall x.\) \(\begin{array}{l}Pt \Leftrightarrow \sqrt {{x^2} – 12x + 36} = 7\\ \Leftrightarrow \sqrt {{{\left( {x – 6} \right)}^2}} = 7\\ \Leftrightarrow \left| {x – 6} \right| = 7\\ \Leftrightarrow \left[ \begin{array}{l}x – 6 = 7\\x – 6 = – 7\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 13\\x = – 1\end{array} \right..\end{array}\) Vậy \(x \in \left\{ { – 1;\;13} \right\}.\)
|
\(b)\;\sqrt {2x + 1} + 1 = x\) Điều kiện: \(2x + 1 \ge 0 \Leftrightarrow x \Leftrightarrow – \dfrac{1}{2}.\) \(\begin{array}{l}Pt \Leftrightarrow \sqrt {2x + 1} = x – 1\\ \Leftrightarrow \left\{ \begin{array}{l}x – 1 \ge 0\\2x + 1 = {\left( {x – 1} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\2x + 1 = {x^2} – 2x + 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\{x^2} – 4x = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\\left[ \begin{array}{l}x = 0\\x = 4\end{array} \right.\end{array} \right. \Leftrightarrow x = 4.\end{array}\) Vậy \(x = 4.\) \(d)\;3\sqrt {27x} – 2\sqrt {12x} – 5 = 10\) Điều kiện: \(x \ge 0.\) \(\begin{array}{l}Pt \Leftrightarrow 3\sqrt {{3^2}.3x} – 2\sqrt {{2^2}.3x} – 5 = 10\\ \Leftrightarrow 9\sqrt {3x} – 6\sqrt {3x} – 15 = 0\\ \Leftrightarrow 3\sqrt {3x} = 15\\ \Leftrightarrow 27x = 225\\ \Leftrightarrow x = \dfrac{{25}}{3}\;\;\left( {tm} \right).\end{array}\) Vậy \(x = \dfrac{{25}}{3}.\) \(f)\;\sqrt {x + 3 + 4\sqrt {x – 1} } + \sqrt {x + 8 – 6\sqrt {x – 1} } = 5\) \(\begin{array}{l} \Leftrightarrow \sqrt {x – 1 + 4\sqrt {x – 1} + 4} + \sqrt {x – 1 – 6\sqrt {x – 1} + 9} = 5\\ \Leftrightarrow \sqrt {{{\left( {\sqrt {x – 1} + 2} \right)}^2}} + \sqrt {{{\left( {\sqrt {x – 1} – 3} \right)}^2}} = 5\end{array}\) Điều kiện: \(x – 1 \ge 0 \Leftrightarrow x \ge 1.\) \(\begin{array}{l}PT \Leftrightarrow \left| {\sqrt {x – 1} + 2} \right| + \left| {\sqrt {x – 1} – 3} \right| = 5\\ \Leftrightarrow \sqrt {x – 1} + 2 + \left| {\sqrt {x – 1} – 3} \right| = 5\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt {x – 1} + \sqrt {x – 1} – 3 = 3\;\;\;\;\;\left( {khi\;\;\sqrt {x – 1} – 3 \ge 0} \right)\\\sqrt {x – 1} – \sqrt {x – 1} + 3 = 2\;\;\;\;\;\left( {khi\;\;\sqrt {x – 1} – 3 < 0} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2\sqrt {x – 1} = 6\;\;\;\;\left( {khi\;\;x \ge 9} \right)\\3 = 2\;\;\left( {VN} \right)\end{array} \right.\\ \Leftrightarrow \sqrt {x – 1} = 3\\ \Leftrightarrow x – 1 = 9\\ \Leftrightarrow x = 10\;\;\left( {tm} \right)\end{array}\) Vậy \(x = 10.\) |