=>x(3x^2-13x+4)=0
=>x(3x^2-12x-x+4)=0
=>x(x-4)(3x-1)=0
=>\(x\in\left\{0;4;\dfrac{1}{3}\right\}\)
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=>x(3x^2-13x+4)=0
=>x(3x^2-12x-x+4)=0
=>x(x-4)(3x-1)=0
=>\(x\in\left\{0;4;\dfrac{1}{3}\right\}\)
Phan tich da thuc thanh nhan tu : x2 - 4x -y2+4
4x^2-36x+56 phan tich da thuc thanh nhan tu
phan tich da thuc thanh nhan tu
16x(x+1)-(4x+3)(4x-3)
thanks
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