a: \(A=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x^2+3x+9\right)}\cdot\dfrac{x^2+3x+9}{x+3}+\dfrac{x}{x+3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3}{x-3}+\dfrac{x}{x+3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x+9+x^2-3x+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x-3}\)
b: \(B=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
Thay x=-1/3 vào B, ta được:
\(B=\dfrac{-1}{3}:\left(3\cdot\dfrac{-1}{3}-1\right)=\dfrac{-1}{3}:\left(-2\right)=\dfrac{1}{6}\)
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