4x²-[2x-(x²-2x+3)-2(x²-3x)]
=4x²-2x+(x²-2x+3)+2(x²-3x)]
=4x²-2x+x²-2x+3+2x²-6x
=7x²-10x+3
4x²-[2x-(x²-2x+3)-2(x²-3x)] 計(jì)算
4x²-[2x-(x²-2x+3)-2(x²-3x)] 計(jì)算
數(shù)學(xué)人氣：280 ℃時(shí)間：2020-01-28 17:37:37
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