Đáp án:
g) x=3
Giải thích các bước giải:
\(\begin{array}{l}
d)DK:\left[ \begin{array}{l}
x \ge 0\\
x \le - 3
\end{array} \right.\\
- {x^2} - 3x + 10 = 3\sqrt {{x^2} + 3x} \\
\to {x^2} + 3x - 10 + 3\sqrt {{x^2} + 3x} = 0\\
Đặt:\sqrt {{x^2} + 3x} = t\left( {t \ge 0} \right)\\
\to {x^2} + 3x = {t^2}\\
Pt \to {t^2} - 10 + 3t = 0\\
\to \left[ \begin{array}{l}
t = 2\\
t = - 5\left( l \right)
\end{array} \right.\\
\to {x^2} + 3x = 4\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 4
\end{array} \right.\left( {TM} \right)\\
g)DK:4 \ge x \ge 2\\
Đặt:\sqrt {x - 2} + \sqrt {4 - x} = t\left( {t \ge 0} \right)\\
\to x - 2 + 2\sqrt { - {x^2} + 6x - 8} + 4 - x = {t^2}\\
\to 2 + 2\sqrt { - \left( {{x^2} - 6x + 8} \right)} = {t^2}\\
\to 2\sqrt { - \left( {{x^2} - 6x + 8} \right)} = {t^2} - 2\\
\to \sqrt { - \left( {{x^2} - 6x + 8} \right)} = \dfrac{{{t^2} - 2}}{2}\\
\to - \left( {{x^2} - 6x + 8} \right) = \dfrac{{{t^4} - 4{t^2} + 4}}{4}\\
\to {x^2} - 6x + 8 = - \dfrac{{{t^4} - 4{t^2} + 4}}{4}\\
\to {x^2} - 6x + 11 = - \dfrac{{{t^4} - 4{t^2} + 4}}{4} + 3\\
Pt \to t = - \dfrac{{{t^4} - 4{t^2} + 4}}{4} + 3\\
\to t - 3 = - \dfrac{{{t^4} - 4{t^2} + 4}}{4}\\
\to 3 - t = \dfrac{{{t^4} - 4{t^2} + 4}}{4}\\
\to 12 - 4t = {t^4} - 4{t^2} + 4\\
\to {t^4} - 4{t^2} + 4t - 8 = 0\\
\to {t^4} - 2{t^3} + 2{t^3} - 4{t^2} + 4t - 8 = 0\\
\to {t^3}\left( {t - 2} \right) + 2{t^2}\left( {t - 2} \right) + 4\left( {t - 2} \right) = 0\\
\to \left( {t - 2} \right)\left( {{t^3} + 2{t^2} + 4} \right) = 0\\
\to \left[ \begin{array}{l}
t = 2\\
t = - 2,594313016\left( l \right)
\end{array} \right.\\
\to - \left( {{x^2} - 6x + 8} \right) = 1\\
\to {x^2} - 6x + 9 = 0\\
\to {\left( {x - 3} \right)^2} = 0\\
\to x = 3
\end{array}\)
\(\begin{array}{l}
e)DK:x \ge \dfrac{2}{3}\\
\sqrt {5x - 1} = \sqrt {3x - 2} + \sqrt {2x + 2} \\
\to 5x - 1 = 3x - 2 + 2\sqrt {\left( {3x - 2} \right)\left( {2x + 2} \right)} + 2x + 2\\
\to - 1 = 2\sqrt {\left( {3x - 2} \right)\left( {2x + 2} \right)} \left( {vô lý} \right)
\end{array}\)
⇒ Phương trình vô nghiệm
\(\begin{array}{l}
f)DK:x \ge 3\\
\sqrt {x - 3} = 5 + \sqrt {x + 2} \\
\to x - 3 = 25 + 10\sqrt {x + 2} + x + 2\\
\to - 30 = 10\sqrt {x + 2} \left( {vô lý} \right)
\end{array}\)
⇒ Phương trình vô nghiệm
