a: \(P=\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right)\cdot\dfrac{9x^2-6x+1}{6x^3+10x}\)
\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x+1\right)\left(3x-1\right)}\cdot\dfrac{\left(3x-1\right)^2}{2x\left(3x^2+5\right)}\)
\(=\dfrac{-x\left(3x^2+5\right)}{\left(3x+1\right)}\cdot\dfrac{3x-1}{2x\left(3x^2+5\right)}=\dfrac{-3x+1}{2\left(3x+1\right)}\)
b: |3x+1|=2
=>3x+1=2 hoặc 3x+1=-2
=>x=-1
Thay x=-1 vào P, ta được:
\(P=\dfrac{-3\cdot\left(-1\right)+1}{2\left(3\cdot-1+1\right)}=\dfrac{5}{2\left(-2+1\right)}=\dfrac{5}{-2}=\dfrac{-5}{2}\)
c: Để P là số nguyên thì -3x+1 chiahết cho 6x+2
=>-6x+2 chia hết cho 6x+2
=>\(6x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{-1\right\}\)