Giải các bất phương trình sau:
a) $\frac{5x+2\sqrt{3-x}}{4}$ > $\frac{x}{4}$ – $\frac{4-3\sqrt{3-x}}{6}$
b) $\frac{x-2}{x-1}$ > $\frac{-3}{x-1}$
Giải các bất phương trình sau:
a) $\frac{5x+2\sqrt{3-x}}{4}$ > $\frac{x}{4}$ – $\frac{4-3\sqrt{3-x}}{6}$
b) $\frac{x-2}{x-1}$ > $\frac{-3}{x-1}$
Đáp án:
$\begin{array}{l}
a)Dkxd:x \le 3\\
\dfrac{{5x + 2\sqrt {3 – x} }}{4} > \dfrac{x}{4} – \dfrac{{4 – 3\sqrt {3 – x} }}{6}\\
\Rightarrow \dfrac{{3.\left( {5x + 2\sqrt {3 – x} } \right)}}{{12}} > \dfrac{{3x – 8 – 6\sqrt {3 – x} }}{{12}}\\
\Rightarrow 15x + 6\sqrt {3 – x} > 3x – 8 – 6\sqrt {3 – x} \\
\Rightarrow 15x – 3x + 8 + 12\sqrt {3 – x} > 0\\
\Rightarrow 12x + 8 + 12\sqrt {3 – x} > 0\\
\Rightarrow 3x + 2 + 3\sqrt {3 – x} > 0\\
\Rightarrow 3\sqrt {3 – x} > – 3x – 2\left( {x < – \dfrac{2}{3}\left( {tm} \right)} \right)\\
+ Khi: – \dfrac{2}{3} < x \le 3\\
\Rightarrow 9\left( {3 – x} \right) > 9{x^2} + 12x + 4\\
\Rightarrow 9{x^2} + 21x – 23 < 0\\
\Rightarrow \dfrac{{ – 7 – \sqrt {141} }}{6} < x < \dfrac{{ – 7 + \sqrt {141} }}{6}\\
Do: – \dfrac{2}{3} < x \le 3 \Rightarrow \dfrac{{ – 2}}{3} < x < \dfrac{{ – 7 + \sqrt {141} }}{6}\\
Vậy\,x < \dfrac{{ – 7 + \sqrt {141} }}{6}\\
b)Dkxd:x \ne 1\\
\dfrac{{x – 2}}{{x – 1}} > \dfrac{{ – 3}}{{x – 1}}\\
\Rightarrow \dfrac{{x – 2}}{{x – 1}} + \dfrac{3}{{x – 1}} > 0\\
\Rightarrow \dfrac{{x + 1}}{{x – 1}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 1\\
x < – 1
\end{array} \right.\\
Vậy\,x > 1\,hoặc\,x < – 1
\end{array}$