Question

Answer 1

a) \(\dfrac{A}{x-2}=\dfrac{5x^2+13x+6}{x^2-4}\)

\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{\left(5x+3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{5x+3}{x-2}\Leftrightarrow A=5x+3\)

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

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

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

Answer 2

a. \(\dfrac{A}{x-2}=\dfrac{5x^2+13x+6}{x^2-4}\)

<=> \(\dfrac{A\left(x+2\right)}{x^2-4}=\dfrac{5x^2+13x+6}{x^2-4}\)

<=> A(x + 2) = 5x² + 10x + 3x + 6

<=> A(x + 2) = 5x(x + 2) - 3(x + 2)

<=> A(x + 2) = (5x - 3)(x + 2)

<=> A = \(\dfrac{\left(5x-3\right)\left(x+2\right)}{x+2}\)

<=> A = 5x - 3

Answer 3

b. \(\dfrac{x-3}{x^2+x+1}=\dfrac{P}{x^3-1}\)

<=> \(\dfrac{\left(x-3\right)\left(x-1\right)}{x^3-1}=\dfrac{P}{x^3-1}\)

<=> P = (x - 3)(x - 1)

Answer 4

a/ Theo đề bài, ta có:

              A(x² - 4) = (x - 2)(5x² + 13x + 6)

⇔ A(x + 2)(x - 2) = (x - 2)(5x² + 13x + 6)

⇔           A(x + 2) = 5x² + 13x + 6

⇔                     A = \(\dfrac{\text{5x^2 + 13x + 6}}{\text{x + 2}}\)

b/ Theo đề bài, ta có:

                       (x - 3)(x³ - 1) = P(x² + x + 1)

⇔   (x - 3)(x - 1)(x² + x + 1) = P(x² + x + 1)

⇔                                     P = (x - 3)(x - 1)