Question

Bài 1: Tìm phân thức, đa thức thỏa mãn điều kiện sau

a)4x²-3x-7/A=4x-7/2x+3

b)a+b/a³+b³=1/B

c)(x²+1).C=2x³+3

d)(x³-1)=(x+1).P

e)x⁴-1=(x+1).Q

Answer 1

a) Ta có: \(\dfrac{4x^2-3x-7}{A}=\dfrac{4x-7}{2x+3}\)

\(\Leftrightarrow A=\dfrac{\left(2x+3\right)\left(4x^2-3x-7\right)}{4x-7}\)

\(\Leftrightarrow A=\dfrac{\left(2x+3\right)\left(4x-7\right)\left(x+1\right)}{4x-7}\)

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

\(\Leftrightarrow A=2x^2+5x+3\)

Answer 2

b) Ta có: \(\dfrac{1}{B}=\dfrac{a+b}{a^3+b^3}\)

\(\Leftrightarrow\dfrac{1}{B}=\dfrac{a+b}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\dfrac{1}{a^2-ab+b^2}\)

hay \(B=a^2-ab+b^2\)

Answer 3

c) Ta có: \(\left(x^2+1\right)\cdot C=2x^3+3\)

\(\Leftrightarrow C=\dfrac{2x^3+3}{x^2+1}\)

Answer 4

e) Ta có: \(Q\left(x+1\right)=x^4-1\)

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

\(\Leftrightarrow Q=\left(x-1\right)\left(x^2+1\right)\)

\(\Leftrightarrow Q=x^3-x^2+x-1\)