Bài 2:
a: Ta có: \(\dfrac{a+2}{a-2}=\dfrac{\left(a+2\right)^2}{A}\)
\(\Leftrightarrow A=\dfrac{\left(a+2\right)^2\cdot\left(a-2\right)}{a+2}\)
\(\Leftrightarrow A=a^2-4\)
b: Ta có: \(\dfrac{A}{2x-1}=\dfrac{6x^3+3x^2}{4x^2-1}\)
\(\Leftrightarrow\dfrac{A}{2x-1}=\dfrac{3x^2\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(\Leftrightarrow\dfrac{A}{2x-1}=\dfrac{3x^2}{2x-1}\)
hay \(A=3x^2\)
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