a) \(3 - 4x\left( {25 - 2x} \right) = 8{x^2} + x - 300\)
\(\Leftrightarrow 3 - 100x + 8{x^2} = 8{x^2} + x - 300\)
\(\Leftrightarrow - 101x = - 303\)
\( \Leftrightarrow x = \left( { - 303} \right):\left( { - 101} \right)\)
\(\Leftrightarrow x = 3\)
Vậy phương trình có nghiệm \(x = 3\) .
b) \(\dfrac{{2\left( {1 - 3x} \right)}}{5} - \dfrac{{2 + 3x}}{{10}} = 7\)\(\, - \dfrac{{3\left( {2x + 1} \right)}}{4}\)
\(\Leftrightarrow \dfrac{{4.2\left( {1 - 3x} \right)}}{{20}} - \dfrac{{2.(2 + 3x)}}{{20}} = \dfrac{{140}}{{20}}\)\(\,- \dfrac{{5.3\left( {2x + 1} \right)}}{{20}}\)
\(\Leftrightarrow 8\left( {1 - 3x} \right) - 2\left( {2 + 3x} \right) = 140 \) \(- 15\left( {2x + 1} \right)\)
\(\Leftrightarrow 8 - 24x - 4 - 6x = 140 - 30x - 15\)
\(\Leftrightarrow - 30x + 4 = 125 - 30x\)
\(\Leftrightarrow -121 = 0x\) (Vô lí)
Vậy phương trình vô nghiệm.
c) \(\dfrac{{5x + 2}}{6} - \dfrac{{8x - 1}}{3} = \dfrac{{4x + 2}}{5} - 5\)
\( \Leftrightarrow \dfrac{{5.(5x + 2)}}{{30}} - \dfrac{{10.(8x - 1)}}{{30}}\)\(\, = \dfrac{{6.(4x + 2)}}{{30}} - \dfrac{{150}}{{30}}\)
\(\Leftrightarrow 5\left( {5x + 2} \right) - 10\left( {8x - 1} \right) \) \(= 6\left( {4x + 2} \right) - 150\)
\(\Leftrightarrow 25x + 10 - 80x + 10\) \( = 24x + 12 - 150\)
\(\Leftrightarrow - 55x + 20 = 24x - 138\)
\(\Leftrightarrow - 79x = - 158\)
\( \Leftrightarrow x = \left( { - 158} \right):\left( { - 79} \right)\)
\(\Leftrightarrow x = 2\)
Vậy phương có nghiệm \(x = 2\).
d) \(\dfrac{{3x + 2}}{2} - \dfrac{{3x + 1}}{6} = 2x + \dfrac{5}{3}\)
\( \Leftrightarrow \dfrac{{3.(3x + 2)}}{6} - \dfrac{{3x + 1}}{6} = \dfrac{{6.2x}}{6}\)\(\, + \dfrac{{5.2}}{6}\)
\(\Leftrightarrow 3\left( {3x + 2} \right) - \left( {3x + 1} \right) = 12x + 10\)
\(\Leftrightarrow 9x + 6 - 3x - 1 = 12x + 10\)
\(\Leftrightarrow 6x + 5 = 12x + 10\)
\( \Leftrightarrow - 6x = 5\)
\(\Leftrightarrow x =\dfrac{{ - 5}}{6}\)
Vậy phương trình có nghiệm \(x =\dfrac{{ - 5}}{6}\).
