\(a,A\left(x\right)=-3x^3+2x^2-6+5x+4x^3-2x^2-4-4x\\ =\left(-3x^3+4x^3\right)+\left(2x^2-2x^2\right)+\left(5x-4x\right)+\left(-6-4\right)\\ =x^3+0+x-10\\ =x^3+x-10\)

Bậc của đa thức \(3\)

Hệ số cao nhất là \(1\)

\(b,B\left(x\right)=A\left(x\right).\left(x-1\right)=\left(x^3+x-10\right)\left(x-1\right)\\ =x^3.x+x.x-10x-x^3-x+10\\ =x^4+x^2-x^3-x-10x+10\\ =x^4-x^3+x^2-11x+10\)

Thay \(x=2\) vào \(B\left(x\right)\)

\(=2^4-2^3+2^2-11.2+10\\ =0\) 

Vậy tại \(x=2\) thì \(B\left(x\right)=0\)

\(a,A\left(x\right)=-3x^3+2x^2-6+5x+4x^3-2x^2-4-4x\\ =\left(-3x^3+4x^3\right)+\left(2x^2-2x^2\right)+\left(5x-4x\right)+\left(-6-4\right)\\ =x^3+0+x-10\\ =x^3+x-10\)

Bậc của đa thức : \(3\)

Hệ số cao nhất ứng với hệ số của số mũ cao nhất : \(1\)

b, \(B\left(x\right)=A\left(x\right).\left(x-1\right)\\ =\left(x^3+x-10\right)\left(x-1\right)\\ =x^3.x+x.x-10x-x^3-x+10\\ =x^4+x^2-x^3-10x-x+10\\ =x^4-x^3+x^2-11x+10\)

\(B\left(2\right)=2^4-2^3+2^2-11.2+10=0\)

a: \(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)

b: \(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)

a)\(f\left(x\right)+g\left(x\right)=\left(x^3+x^3\right)-2x^2+\left(3x+x\right)+\left(1-1\right)\)

\(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)

b)\(f\left(x\right)-g\left(x\right)=x^3-2x^2+3x+1-x^3-x+1\)

\(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)

\(a,A\left(x\right)=2x^3-3x^2+2x+1\\ B\left(x\right)=3x^3+2x^2-x-5\\ b,A\left(x\right)+B\left(x\right)=\left(2x^3+3x^3\right)+\left(2x^2-3x^2\right)+\left(2x-x\right)+\left(1-5\right)=5x^3-x^2+x-4\\ A\left(x\right)-B\left(x\right)=\left(2x^3-3x^3\right)+\left(-3x^2-2x^2\right)+\left(2x+x\right)+\left(1-5\right)=-x^3-5x^2+3x-4\)

\(a,\Leftrightarrow4x^3-2x^2+a=\left(2x-3\right).a\left(x\right)\)

Thay \(x=\dfrac{3}{2}\Leftrightarrow4.\dfrac{27}{8}-2.\dfrac{9}{4}+a=0\)

\(\Leftrightarrow\dfrac{27}{2}-\dfrac{9}{2}+a=0\\ \Leftrightarrow a=-9\)

\(b,\Leftrightarrow3x^3+2x^2+x+a=\left(x+1\right).b\left(x\right)+2\)

Thay \(x=-1\Leftrightarrow-3+2-1+a=2\Leftrightarrow a=4\)

a)

\(A\left(x\right)=3x^3+3x^2+2x-1\)

Bậc của A(x) là 3

Hệ số tự do A(x) là -1

Hệ số cao nhất của A(x) là 3

Tại A(-2)

\(A=3.\left(-2\right)^3+3.\left(-2\right)^2+2.\left(-2\right)-1\)

\(=-17\)

b)

\(B\left(x\right)=5x^4+6x-2x^2+4-5x^4-5x\)

\(=\left(5x^4-5x^4\right)+\left(-2x^2\right)+\left(6x-5x\right)+4\)

\(=-2x^2+x+4\)

c)

\(A\left(x\right)-B\left(x\right)=3x^3+3x^2+2x-1-\left(-2x^2+x+4\right)\)

                              \(=3x^3+3x^2+2x-1+2x^2-x-4\)

                              \(=3x^3+\left(3x^2+2x^2\right)+\left(2x-x\right)+\left(-1-4\right)\)

                              \(=3x^3+5x^2+x-5\)

d)

\(C\left(x\right)-2.\left(-2x^2+x+4\right)=3x^3+3x^2+2x-1\)

\(C\left(x\right)=3x^3+3x^2+2x-1+2.\left(-2x^2+x+4\right)\)

\(C\left(x\right)=3x^3+3x^2+2x-1-4x^2+2x+8\)

\(C\left(x\right)=3x^3+\left(3x^2-4x^2\right)+\left(2x+2x\right)+\left(-1+8\right)\)

\(C\left(x\right)=3x^3-x^2+4x+7\)

chúc bạn học giỏi

`a,A(x)=2x^3+2x-3x^2+11`

`=2x^3-3x^2+2x+11`

`B(x)=2^2+3x^3-x-5`

`=3x^3+2x^2-x-5`

`b, A(x)+B(x)=(2x^3-3x^2+2x+11)+(3x^3+2x^2-x-5)`

`=2x^3-3x^2+2x+11+3x^3+2x^2-x-5`

`=(2x^3+3x^3)+(-3x^2+2x^2)+(2x-x)+(11-5)`

`=5x^3 -x^2 +x+6`

`c,A(x)-B(x)=(2x^3-3x^2+2x+11)-(3x^3+2x^2-x-5)`

`=2x^3-3x^2+2x+11- 3x^3 -2x^2+x+5`

`=(2x^3-3x^3)+(-3x^2-2x^2)+(2x+x)+(11+5)`

`=-x^3 -5x^2+3x+16`

a/\(A\left(x\right)=2x^3+2x-3x^2+11\)
\(=2x^3-3x^2+2x+11\)
\(B\left(x\right)=2x^2+3x^3-x-5\)
\(=3x^3+2x^2-x-5\)
b/\(A\left(x\right)+B\left(x\right)=\left(2x^3-3x^2+2x+11\right)+\left(3x^3+2x^2-x-5\right)\)
\(=2x^3-3x^2+2x+11+3x^3+2x^2-x-5\)
\(=\left(2x^3+3x^3\right)-\left(3x^2-2x^2\right)+\left(2x-x\right)+\left(11-5\right)\)
\(=5x^3-x^2+x+6\)
c/\(A\left(x\right)+B\left(x\right)=\left(2x^3-3x^2+2x+11\right)-\left(3x^3+2x^2-x-5\right)\)
\(=2x^3-3x^2+2x+11-3x^3-2x^2+x+5\)
\(=\left(2x^3-3x^3\right)-\left(3x^2+2x^2\right)+\left(2x+x\right)+\left(11+5\right)\)
\(=-x^3-5x^2+3x+16\)
#DarkPegasus

\(\text{a ) A ( x ) = 2 x 3 − 3 x 2 + 2 x + 1}\)

\(\text{B ( x ) = 3 x 3 + 2 x 2 − x − 5 }\)

\(\text{b ) A ( x ) + B ( x ) = 2 x 3 − 3 x 2 + 2 x + 1 + 3 x 3 + 2 x 2 − x − 5}\)

\(\text{= ( 2 x 3 + 3 x 3 ) + ( − 3 x 2 + 2 x 2 ) + ( 2 x − x ) + ( 1 − 5 )}\)

\(\text{= 5 x 3 − x 2 + x − 4 }\)
 

giúp em  bài với ạ,em cảm ơn, em đang vội ạ

\(a,=-15x^3+10x^4+20x^2\\ b,=2x^3+2x^2+4x-x^2-x-2=2x^3+x^2+3x-2\)