Ta có :
\(B=\frac{x^3}{x+1}+\frac{x^2}{x-3}+\frac{1}{x+1}-\frac{9}{x-3}\)
\(=\frac{x^3+1}{x+1}+\frac{x^2-9}{x-3}\)
\(=\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}+\frac{\left(x-3\right)\left(x+3\right)}{x-3}\)
\(=x^2-x+1+x+3\)
\(=x^2+4\)
Với mọi x ta có :
\(x^2\ge0\)
\(\Leftrightarrow x^2+4\ge4\)
\(\Leftrightarrow B\ge4\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
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