Đáp án:
f) x=-2
Giải thích các bước giải:
\(\begin{array}{l}
d)DK:x \ne \left\{ { - 2;3} \right\}\\
\dfrac{{\left( {x + 2} \right)\left( {3 - x} \right) + x\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {3 - x} \right)}} = \dfrac{{5x + 2\left( {3 - x} \right)}}{{\left( {x + 2} \right)\left( {3 - x} \right)}}\\
\to - {x^2} + x + 6 + {x^2} + 2x = 5x + 6 - 2x\\
\to 6 = 6\left( {ld} \right)
\end{array}\)
⇒ Phương trình có vô số nghiệm với \(x \ne \left\{ { - 2;3} \right\}\)
\(\begin{array}{l}
e)DK:x \ne \pm 1\\
\dfrac{{\left( {2x + 1} \right)\left( {x + 1} \right) - 5{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to 2{x^2} + 3x + 1 - 5\left( {{x^2} - 2x + 1} \right) = 0\\
\to - 3{x^2} + 13x - 4 = 0\\
\to \left( {4 - x} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = \dfrac{1}{3}
\end{array} \right.\\
f)DK:x \ne \left\{ { - \dfrac{7}{2}; - 3;3} \right\}\\
\dfrac{{13 - x + 3}}{{\left( {x - 3} \right)\left( {2x + 7} \right)}} = \dfrac{{6\left( {2x + 7} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)\left( {2x + 7} \right)}}\\
\to \dfrac{{\left( {16 - x} \right)\left( {x + 3} \right) - 12x - 42}}{{\left( {x - 3} \right)\left( {x + 3} \right)\left( {2x + 7} \right)}} = 0\\
\to - {x^2} + 13x + 48 - 12x - 42 = 0\\
\to - {x^2} + x + 6 = 0\\
\to \left[ \begin{array}{l}
x = 3\left( l \right)\\
x = - 2
\end{array} \right.
\end{array}\)
