2 bài tập Giải phương trình (có lời giải)

a) Điều kiện: $x \ge - \frac{1}{2}$

$\sqrt {18x + 9} + \sqrt {x + 3} - \frac{1}{2}\sqrt {4x + 12} = 9$

$\Leftrightarrow 3\sqrt {2x + 1} + \sqrt {x + 3} - \sqrt {x + 3} = 9$

$\Leftrightarrow \sqrt {2x + 1} = 3 \Leftrightarrow 2x + 1 = 9 \Leftrightarrow x = 4$ (nhận)

Vậy phương trình có nghiệm $x = 4$

a) Điều kiện: $x \ge 2$

$\sqrt {25x - 50} - \sqrt {16x - 32} - \sqrt {9x - 18} = 12 - 4\sqrt {x - 2}$

$$\Leftrightarrow 5\sqrt {x - 2} - 4\sqrt {x - 2} - 3\sqrt {x - 2} = 12 - 4\sqrt {x - 2}$$

$\Leftrightarrow \sqrt {x - 2} = 6 \Leftrightarrow x - 2 = 36 \Leftrightarrow x = 38$ (nhận)

Vậy phương trình có nghiệm $x = 38$

a) $2\sqrt {9x - 27} - \frac{1}{5}\sqrt {25x - 75} - \frac{1}{7}\sqrt {49x - 147} = 20$

ĐKXĐ: $x \ge 3$

$\begin{array}{l}2\sqrt {9x - 27} - \frac{1}{5}\sqrt {25x - 75} - \frac{1}{7}\sqrt {49x - 147} = 20\\2\sqrt {9\left( {x - 3} \right)} - \frac{1}{5}\sqrt {25\left( {x - 3} \right)} - \frac{1}{7}\sqrt {49\left( {x - 3} \right)} = 20\\6\sqrt {x - 3} - \sqrt {x - 3} - \sqrt {x - 3} = 20 \Leftrightarrow 4\sqrt {x - 3} = 20\\\sqrt {x - 3} = 5 \Leftrightarrow x - 3 = 25\end{array}$

$x = 28$ (TMĐK)

Vậy $x = 28.$

b)$\frac{{3\sqrt x - 5}}{2} - \frac{{2\sqrt x - 7}}{3} + 1 = \sqrt x$

ĐKXĐ: $x \ge 0$

$\begin{array}{l}\frac{{3\sqrt x - 5}}{2} - \frac{{2\sqrt x - 7}}{3} + 1 = \sqrt x \\\frac{{\left( {3\sqrt x - 5} \right).3}}{6} - \frac{{\left( {2\sqrt x - 7} \right).2}}{6} + \frac{6}{6} = \frac{6}{6}\sqrt x \\9\sqrt x - 15 - 4\sqrt x + 14 + 6 = 6\sqrt x \\9\sqrt x - 4\sqrt x - 6\sqrt x = 15 - 14 - 6\\ - \sqrt x = - 5\end{array}$

$\sqrt x = 5$

$x = 25$ (TMĐK)

Vậy $x = 25.$

c)$\sqrt {9{x^2} + 45} - \frac{1}{{12}}\sqrt {16{x^2} + 80} + 3\sqrt {\frac{{{x^2} + 5}}{{16}}} - \frac{1}{4}\sqrt {\frac{{25{x^2} + 125}}{9}} = 9;$

$$\begin{array}{l}\sqrt {9{x^2} + 45} - \frac{1}{{12}}\sqrt {16{x^2} + 80} + 3\sqrt {\frac{{{x^2} + 5}}{{16}}} - \frac{1}{4}\sqrt {\frac{{25{x^2} + 125}}{9}} = 9\\\sqrt {9\left( {{x^2} + 5} \right)} - \frac{1}{{12}}\sqrt {16\left( {{x^2} + 5} \right)} + 3\frac{{\sqrt {{x^2} + 5} }}{{\sqrt {16} }} - \frac{1}{4}\frac{{\sqrt {25\left( {{x^2} + 5} \right)} }}{{\sqrt 9 }} = 9\\3\sqrt {{x^2} + 5} - \frac{1}{3}\sqrt {{x^2} + 5} + \frac{3}{4}\sqrt {{x^2} + 5} - \frac{5}{{12}}\sqrt {{x^2} + 5} = 9 \Leftrightarrow 3\sqrt {{x^2} + 5} = 9\\\sqrt {{x^2} + 5} = 3\\{x^2} + 5 = 9\\{x^2} = 4\end{array}$$

$x = - 2$ hoặc $x = 2.$

Vậy $x = - 2$ hoặc $x = 2.$

d) $\sqrt {4{,}5x} + \sqrt {50x} - \sqrt {32x} + \sqrt {72x} - 5\sqrt {\frac{x}{2}} - 12 = 0.$

ĐKXĐ: $x \ge 0$

$\begin{array}{l}\sqrt {4{,}5x} + \sqrt {50x} - \sqrt {32x} + \sqrt {72x} - 5\sqrt {\frac{x}{2}} - 12 = 0\\\sqrt {\frac{{9x}}{2}} + \sqrt {{5^2}.2x} - \sqrt {{4^2}.2x} + \sqrt {{6^2}.2x} - 5\sqrt {\frac{x}{2}} - 12 = 0\\\frac{3}{2}\sqrt {2x} + 5\sqrt {2x} - 4\sqrt {2x} + 6\sqrt {2x} - \frac{5}{2}\sqrt {2x} = 12 \Leftrightarrow 6\sqrt {2x} = 12\\\sqrt {2x} = 2\\2x = 4\end{array}$

$x = 2$ (TMĐK)

Vậy $x = 2.$