Question

giúp mình nha mk cảm ơn

plssss đang gấp lắm nhaaaa!

Answer 1

a: \(A=2x^3-3x^2+4x+5\)

\(\dfrac{A}{B}=\dfrac{2x^3-3x^2+4x+5}{2x+1}\)

\(=\dfrac{2x^3+x^2-4x^2-2x+6x+3+2}{2x+1}=x^2-2x+3+\dfrac{2}{2x+1}\)

b: \(\dfrac{A\left(x\right)}{B\left(x\right)}=\dfrac{2x^3-3x^2+4x+a}{2x+1}\)

\(=\dfrac{2x^3+x^2-4x^2-2x+6x+3+a-3}{2x+1}\)

\(=x^2-2x+3+\dfrac{a-3}{2x+1}\)

Để A chia hết cho B thì a-3=0

=>a=3