计算：(B-1)(1+B)(B²-1),（a-2b+3c)(a+2b-3c)
1个回答
计算：
1.(B-1)(1+B)(B²-1)
=(B-1)(B+1)(B²-1)
=(B²-1)(B²-1)
=(B²-1)²
2.（a-2b+3c)(a+2b-3c)
=[a-(2b-3c)][a+(2b-3c)]
=a² -(2b-3c)²
分解因式：
1.a²x²+16ax +64
=(ax)²+2*8*ax +8²
=(ax+8)²
2.25(x+y)²-16(x-y)²
=[5*(x+y)]²-[4(x-y)]²
=[5(x+y)+4(x-y)]*[(5(x+y)-4(x-y)]
=(9x-y)*(x+9y)
=9x²+81xy-xy-9²
=9x²+80xy-9y²
3.x²-6x+9-Y²
=x²-2*3x+3*3-Y²
=(x-3)²-Y²
=(x-3-y)*(x-3+y)
4.(a²+b²-1）²-4a²b²
=(a²+b²-1）²-(2ab)²
=[(a²+b²-1)+2ab]*[(a²+b²-1)-2ab]
=[(a+b)²-1]*[(a-b)²-1]
求值：
[(x+2y- 3/2)(x-2y+ 3/2)+2y(2y-3)+(2^-2/3)]/(-3x)^-1
=[(x²-(2y- 3/2)²]*(-3x)+[4y²-6y+ (√2)/4]*(-3x)
将X+-1,Y=2007/2006代入即可
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