\(A\left(x\right)⋮B\left(x\right)\)

=>\(x^3-4x^2+ax+30⋮x-5\)

=>\(x^3-5x^2+x^2-5x+\left(a+5\right)x-5a-25+5a+55⋮x-5\)

=>5a+55=0

=>5a=-55

=>a=-11

\(4x^3+ax+b\) chia hết cho x-2

=>\(4x^3-8x^2+8x^2-16x+\left(a+16\right)x-2\left(a+16\right)+2\left(a+16\right)+b\) ⋮x-2

=>2(a+16)+b=0

=>b=-2(a+16)=-2a-32

\(4x^3+ax+b\) chia hết cho x+1

=>\(4x^3+4x^2-4x^2-4x+\left(a+4\right)x+a+4+b-a-4\) ⋮x+1

=>b-a-4=0

=>b=a+4

=>-2a-32=a+4

=>-3a=36

=>a=-12

b=a+4=-12+4=-8

2a-3b

\(=2\cdot\left(-12\right)-3\cdot\left(-8\right)\)

=-24+24=0

sao nhiều bt vại , sao làm hết nổi !!!!! ~ _~

\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)

\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left[\left(a-b\right)+\left(a+c\right)\right]\)

\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left(a-b\right)+bc\left(a+c\right)\)

\(=\left(a-b\right)\left(ab+bc\right)+\left(a+c\right)\left(bc-ac\right)\)

\(=b\left(a-b\right)\left(a+c\right)-c\left(a+c\right)\left(a-b\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+c\right)\)

b: \(\Leftrightarrow2n^2+n-2n-1+3⋮2n+1\)

\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{0;-1;1;-2\right\}\)

a: \(\dfrac{A}{B}=\dfrac{x^3+4x^2+3x+12-19}{x+4}=x^2+3+\dfrac{-19}{x+4}\)

b: Để A chia hết cho B thì \(x+4\in\left\{1;-1;19;-19\right\}\)

=>\(x\in\left\{-3;-5;15;-23\right\}\)