a)\(\left( {2x + 1} \right)\left( {3x - 2} \right) = \left( {5x - 8} \right)\left( {2x + 1} \right)\)
\(\Leftrightarrow\)\( \left( {2x + 1} \right)\left( {3x - 2} \right) - \left( {5x - 8} \right)\left( {2x + 1} \right)\) \( = 0\)
\(\Leftrightarrow \left( {2x + 1} \right)\left( {3x - 2 - 5x + 8} \right) = 0\)
\( \Leftrightarrow \left( {2x + 1} \right)\left( {6- 2x} \right) = 0\)
\( \Leftrightarrow \left[ {\matrix{{2x + 1 = 0} \cr {6 - 2x = 0} \cr} \Leftrightarrow \left[ {\matrix{{x = \dfrac{ - 1} {2}} \cr {x = 3} \cr} } \right.} \right.\)
Vậy phương trình có hai nghiệm \(x = \dfrac{{ - 1}}{2};\; x = {3}\) .
b) \(4{x^2} - 1 = \left( {2x + 1} \right)\left( {3x - 5} \right)\)
\(\Leftrightarrow \left( {2x - 1} \right)\left( {2x + 1} \right) \) \(= \left( {2x + 1} \right)\left( {3x - 5} \right)\)
\(\Leftrightarrow \left( {2x + 1} \right)\left( {2x - 1 - 3x + 5} \right)=0\)
\(\Leftrightarrow \left( {2x + 1} \right)\left( {4 - x} \right) = 0\)
\( \Leftrightarrow \left[ {\matrix{{2x + 1 = 0} \cr {4 - x = 0} \cr} \Leftrightarrow \left[ {\matrix{{x = \dfrac{{ - 1}}{2}} \cr {x = 4} \cr} } \right.} \right.\)
Vậy phương trình có hai nghiệm \(x = \dfrac{{ - 1}}{2};x = 4\)
c) \({\left( {x + 1} \right)^2} = 4\left( {{x^2} - 2x + 1} \right)\)
\(\Leftrightarrow {\left( {x + 1} \right)^2}\) \( = \left[ {2(x - 1} \right){]^2}\)
\( \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {2x - 2} \right)^2} = 0\)
\(\Leftrightarrow \left( {x + 1 - 2x + 2} \right)\left( {x + 1 + 2x - 2} \right) \) \(= 0\)
\(\Leftrightarrow \left( {3 - x} \right)\left( {3x - 1} \right) = 0\)
\(\Leftrightarrow \left[ {\matrix{{3 - x = 0} \cr {3x - 1 = 0} \cr} \Leftrightarrow \left[ {\matrix{{x = 3} \cr {x = \dfrac{1}{3}} \cr} } \right.} \right.\)
Vậy phương trình có hai nghiệm: \( x = 3;\; {x = \dfrac{1}{3}}\)
d) \(2{x^3} + 5{x^2} - 3x = 0\)
\(\Leftrightarrow x\left( {2{x^2} + 5x - 3} \right) = 0\)
\(\Leftrightarrow x(2{x^2} + 6x - x - 3) = 0\)
\(\Leftrightarrow x\left[ {2x\left( {x + 3} \right) - \left( {x + 3} \right)} \right] = 0\)
\(\Leftrightarrow x\left( {x + 3} \right)\left( {2x - 1} \right) = 0\)
\(\Leftrightarrow \left[ {\matrix{{x = 0} \cr {x + 3 = 0} \cr {2x - 1 = 0} \cr} } \right. \Leftrightarrow \left[ {\matrix{{x = 0} \cr {x = - 3} \cr {x =\dfrac{1}{2}} \cr} } \right.\)
Vậy phương trình có ba nghiệm \(x = 0;\; x = -3;\; x =\dfrac{1}{2}\).
