a: \(\Leftrightarrow2x^3-5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2-5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2-6x+x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(2x+1\right)=0\)
hay \(x\in\left\{0;3;-\dfrac{1}{2}\right\}\)
b: \(\Leftrightarrow\left(3x-1\right)\left(x^2+x\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+x-7x+10\right)=0\)
=>3x-1=0
hay x=1/3
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