Question

Bài 3. Tìm đa thức \( M \) biết:
a) \( M \cdot \left( \frac{1}{2}x + 3 \right) = \frac{1}{2}x^3 + 4x^2 + 7,5x + 9 \)

b) \( x^3 - 2x^2 - 15x + 36 = (x^2 - 6x + 9)M \)

c) \( (x^3 + 3x^2 - 1) \cdot M = x^5 + 3x^4 - 2x^3 - 7x^2 + 2 \)

Bài 4. Trong mỗi trường hợp sau, tìm thương \( Q(x) \) và dư \( R(x) \) trong phép chia \( A(x) \) cho \( B(x) \) rồi biểu diễn \( A(x) \) dưới dạng: \( A(x) = B(x) \cdot Q(x) + R(x) \)

a) \( A(x) = 5x^3 + 14x^2 + 12x - 5; \, B(x) = x + 2 \)

b) \( A(x) = -\frac{3}{2}x^3 + 5x^2 - 7x + 11; \, B(x) = -3x + 4 \)

Answer 1

Bài 3

a: \(M\left(\dfrac{1}{2}x+3\right)=\dfrac{1}{2}x^3+4x^2+7,5x+9\)

=>\(M\left(\dfrac{1}{2}x+3\right)=\left(\dfrac{1}{2}x^3+3x^2\right)+\left(x^2+6x\right)+\left(1,5x+9\right)\)

=>\(M\left(\dfrac{1}{2}x+3\right)=x^2\left(\dfrac{1}{2}x+3\right)+2x\left(\dfrac{1}{2}x+3\right)+3\left(\dfrac{1}{2}x+3\right)\)

=>\(M=x^2+2x+3\)

b: \(x^3-2x^2-15x+36=\left(x^2-6x+9\right)\cdot M\)

=>\(M=\dfrac{x^3-2x^2-15x+36}{x^2-6x+9}\)

=>\(M=\dfrac{x^3-6x^2+9x+4x^2-24x+36}{x^2-6x+9}\)

=>M=x+4

c: \(\left(x^3+3x^2-1\right)\cdot M=x^5+3x^4-2x^3-7x^2+2\)

=>\(M=\dfrac{x^5+3x^4-2x^3-7x^2+2}{x^3+3x^2-1}\)

=>\(M=\dfrac{x^5+3x^4-x^2-2x^3-6x^2+2}{x^3+3x^2-1}\)

=>\(M=\dfrac{x^2\left(x^3+3x^2-1\right)-2\left(x^3+3x^2-1\right)}{x^3+3x^2-1}=x^2-2\)

Bài 4

a: \(\dfrac{A\left(x\right)}{B\left(x\right)}=\dfrac{5x^3+14x^2+12x-5}{x+2}\)

\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8-13}{x+2}\)

\(=5x^2+4x+4-\dfrac{13}{x+2}\)

=>Thương là \(Q\left(x\right)=5x^2+4x+4\) và dư là R(x)=-13

Vậy: \(5x^3+14x^2+12x-5=\left(5x^2+4x+4\right)\left(x+2\right)+\left(-13\right)\)

b: \(\dfrac{A\left(x\right)}{B\left(x\right)}=\dfrac{-\dfrac{3}{2}x^3+5x^2-7x+11}{-3x+4}\)

\(=\dfrac{-\dfrac{3}{2}x^3+2x^2+3x^2-4x-3x+4+7}{-3x+4}\)

\(=\dfrac{0,5x^2\left(-3x+4\right)-x\left(-3x+4\right)+\left(-3x+4\right)+7}{-3x+4}\)

\(=0,5x^2-x+1+\dfrac{7}{-3x+4}\)

Thương là \(Q\left(x\right)=0,5x^2-x+1\)

Dư là R(x)=7

Do đó: \(-\dfrac{3}{2}x^3+5x^2-7x+11=\left(-3x+4\right)\left(0,5x^2-x+1\right)+7\)