[(a+b)(a-b)+(a-b)2+4a2(a+1)]÷a
=[a2-b2+a2-2ab+b2+4a3+4a2]÷a
=[4a3+6a2-2ab]÷a
=2(2a2+3a-b).
當(dāng)2a2+3a-b=4時,2(2a2+3a-b)=2×4=8.
若2a2+3a-b=4,求代數(shù)式[(a+b)(a-b)+(a-b)2+4a2(a+1)]÷a的值.
若2a2+3a-b=4,求代數(shù)式[(a+b)(a-b)+(a-b)2+4a2(a+1)]÷a的值.
數(shù)學(xué)人氣:218 ℃時間:2019-08-19 13:29:08
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