Question

giúp mình ik :3

Answer 1

Bài 4:

a: \(2x^3-7x^2+5x+1\) ⋮2x-1

=>\(2x^3-x^2-6x^2+3x+2x-1+2\) ⋮2x-1

=>2⋮2x-1

mà 2x-1 lẻ

nên 2x-1∈{1;-1}

=>2x∈{2;0}

=>x∈{1;0}

b: \(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

\(=\frac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)

\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}=x^2-2x-5\)