求回答,初二数学[(x²+6x+9)÷(x+3)](x²-3x+9)
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求回答,初二数学[(x²+6x+9)÷(x+3)](x²-3x+9)
求回答,初二数学[(x²+6x+9)÷(x+3)](x²-3x+9)
求回答,初二数学[(x²+6x+9)÷(x+3)](x²-3x+9)
原式=(x+63)²÷(x+3)*(x²-3x+9)
=(x+3)(x²-3x+9)
=x³-3x²+9x+3x²-9x+27
=x³+27
[(x²+6x+9)÷(x+3)](x²-3x+9)
=[(x+3)²÷(x+3)](x²-3x+9)
=(x+3)(x²-3x+3²)
=x^3+27