Đáp án:
10) Hệ có vô số nghiệm với \(x \ne \left\{ { - 2;3} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)2\left( {x + \dfrac{3}{5}} \right) = 5 - \dfrac{{13}}{5} - x\\
\to 2x + \dfrac{6}{5} = \dfrac{{12}}{5} - x\\
\to 3x = \dfrac{6}{5}\\
\to x = \dfrac{2}{5}\\
7)5{x^2} - 3x - 5x + 3 = 3{x^2} - 3x - 8x + 8\\
\to 2{x^2} + 3x - 5 = 0\\
\to 2{x^2} - 2x + 5x - 5 = 0\\
\to 2x\left( {x - 1} \right) + 5\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {2x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{5}{2}
\end{array} \right.\\
8){\left( {2x - 1} \right)^2} + \left( {2 - x} \right)\left( {2x - 1} \right) = 0\\
\to \left( {2x - 1} \right)\left( {2x - 1 + 2 - x} \right) = 0\\
\to \left[ \begin{array}{l}
2x - 1 = 0\\
x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - 1
\end{array} \right.\\
9)DK:x \ne \pm 2\\
\dfrac{{\left( {1 - 6x} \right)\left( {x + 2} \right) + \left( {9x + 4} \right)\left( {x - 2} \right) - 3{x^2} + 2x - 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
\to - 6{x^2} - 11x + 2 + 9{x^2} - 14x - 8 - 3{x^2} + 2x - 1 = 0\\
\to - 23x - 7 = 0\\
\to x = - \dfrac{7}{{23}}\\
10)DK:x \ne \left\{ { - 2;3} \right\}\\
\dfrac{{\left( {x + 2} \right)\left( {3 - x} \right) + x\left( {x + 2} \right) - 5x - 2\left( {3 - x} \right)}}{{\left( {x + 2} \right)\left( {3 - x} \right)}} = 0\\
\to - {x^2} + x + 6 + {x^2} + 2x - 5x - 6 + 2x = 0\\
\to 0x = 0\left( {ld} \right)
\end{array}\)
⇒ Hệ có vô số nghiệm với \(x \ne \left\{ { - 2;3} \right\}\)
