4(2p+3q)²-(3p-q)²怎么算. 4a²-(b+c)²怎么算.过程~
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4(2p+3q)²-(3p-q)²怎么算. 4a²-(b+c)²怎么算.过程~
4(2p+3q)²-(3p-q)²怎么算. 4a²-(b+c)²怎么算.过程~
4(2p+3q)²-(3p-q)²怎么算. 4a²-(b+c)²怎么算.过程~
4(2p+3q)²-(3p-q)²
=[2(2p+3q)]²-(3p-q)²
=(4p+6q)²-(3p-q)²
=[(4p+6q)+(3p-q)][(4p+6q)-(3p-q)]
=(7p+5q)(p+7q) 因式分解
=7p²+49pq+5pq+35q²
=7p²+54pq+35q² 求结果
4a²-(b+c)²
=(2a)²-(b+c)²
=(2a+b+c)(2a-b-c) 因式分解
求结果的话直接打开括号
=4a²-(b²+2bc+c²)
=4a²-b²-c²-2bc
4(2p+3q)²-(3p-q)²
=[2(2p+3q)]²-(3p-q)
=(4p+6q+3p-q)(4p+6q-3p+q)
=(7p+5q)(p+7q)
4a²-(b+c)²
=(2a)²-(b+c)²
=(2a+b+c)(2a-b-c)
很高兴为您解答,祝你学习进步!...
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4(2p+3q)²-(3p-q)²
=[2(2p+3q)]²-(3p-q)
=(4p+6q+3p-q)(4p+6q-3p+q)
=(7p+5q)(p+7q)
4a²-(b+c)²
=(2a)²-(b+c)²
=(2a+b+c)(2a-b-c)
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第一个可以全部展开,第二个用平方差公式