Question

1) Tìm đa thức A biết

a) \(\dfrac{a+b}{a^3+b^3}\)=\(\dfrac{1}{a}\)

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

c)(x²+1).a=2x²-3

d) \(\dfrac{a+2}{a-2}\)=\(\dfrac{\left(a+2\right)^2}{a}\)

đ) \(\dfrac{4a^2-12ab+9b^2}{-2ac+3bc}\) giải giùm mình cần gắp cảm ơn nhiều ạ

Answer 1

a: \(\dfrac{1}{A}=\dfrac{a+b}{a^3+b^3}\)

\(\Leftrightarrow A=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)

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

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

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

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

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

nên \(A=\dfrac{2x^2-3}{x^2+1}\)

d: \(\dfrac{a+2}{a-2}=\dfrac{\left(a+2\right)^2}{A}\)

nên \(A=\dfrac{\left(a+2\right)^2\cdot\left(a-2\right)}{a+2}=\left(a+2\right)\left(a-2\right)=a^2-4\)