Đáp án:
B3:
a. \(\dfrac{x}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a.DK:x \ne \pm 2\\
B = \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right) + 5x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} - 3x + 2 + 5x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{x}{{x - 2}}\\
b.\dfrac{A}{B} = \dfrac{{{x^2} + 3}}{{x - 2}}:\dfrac{x}{{x - 2}}\\
= \dfrac{{{x^2} + 3}}{x}\\
= x + \dfrac{3}{x}\\
Do:x > 0\\
Co - si:x + \dfrac{3}{x} \ge 2\sqrt {x.\dfrac{3}{x}} = 2\sqrt 3 \\
\to Min = 2\sqrt 3 \\
\Leftrightarrow x = \dfrac{3}{x}\\
\to {x^2} = 3\\
\to x = \sqrt 3 \\
B5:\\
a.DK:x \ne - 3\\
A = \dfrac{{3\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^2}}} = \dfrac{3}{{x + 3}}\\
DK:x \ne \pm 2\\
B = \dfrac{{x - 2 - x - 2 + 2x + 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{2}{{x - 2}}\\
b.A = B\\
\to \dfrac{3}{{x + 3}} = \dfrac{2}{{x - 2}}\\
\to 3\left( {x - 2} \right) = 2\left( {x + 3} \right)\\
\to 3x - 6 = 2x + 6\\
\to x = 12
\end{array}\)
