Đáp án:
9) \(\left[ \begin{array}{l}
x = \frac{{ - 3 + \sqrt 5 }}{2}\\
x = \frac{{ - 3 - \sqrt 5 }}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
9)DK:x \ne \pm 2\\
Pt \to \frac{{{x^2} + 3x + 2 - 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
\to {x^2} + 3x + 1 = 0\\
\to \left[ \begin{array}{l}
x = \frac{{ - 3 + \sqrt 5 }}{2}\\
x = \frac{{ - 3 - \sqrt 5 }}{2}
\end{array} \right.\left( {TM} \right)\\
11)DK:x \ne \pm 5\\
Pt \to {x^2} + 10x + 25 - {x^2} + 10x - 25 = 20\\
\to 20x = 20\\
\to x = 1\left( {TM} \right)\\
13)DK:x \ne \left\{ {\frac{1}{5};\frac{3}{5};3} \right\}\\
Pt \to \frac{{9 - 15x + 10x - 2}}{{\left( {5x - 1} \right)\left( {3 - 5x} \right)}} = \frac{{ - 4}}{{\left( {5x - 1} \right)\left( {x - 3} \right)}}\\
\to \frac{{\left( {7 - 5x} \right)\left( {x - 3} \right)}}{{\left( {5x - 1} \right)\left( {3 - 5x} \right)\left( {x - 3} \right)}} = \frac{{ - 4\left( {3 - 5x} \right)}}{{\left( {5x - 1} \right)\left( {3 - 5x} \right)\left( {x - 3} \right)}}\\
\to 7x - 21 - 5{x^2} + 15x = - 12 + 20x\\
\to - 5{x^2} + 2x - 9 = 0\\
Do: Δ'= 1 - 4.5.9 = - 179 < 0
\end{array}\)
⇒ Pt vô nghiệm
