1.计算 (1)(-12a^2b^2c)· (-4分之1abc^2)^2 (2)(-a^2)·(-2ab)+3a^2(a

1.计算 (1)(-12a^2b^2c)· (-4分之1abc^2)^2 (2)(-a^2)·(-2ab)+3a^2(ab-3分之1b-1)
1.计算
(1)(-12a^2b^2c)· (-4分之1abc^2)^2
(2)(-a^2)·(-2ab)+3a^2(ab-3分之1b-1)
(3)8x^2-5x(4y-x)+4x(-4x-2分之5y)
(4)(3a^2-2b^2)(a^2+b^2)
2.解方程
1.(2x+5)(x-1)=2(x+4)(x-3)
3.1.已知x+y=m,求(x+y)^3+(2x+2y)^3+(3x+3y)^3的值
蓝域鬼惑 1年前 已收到1个回答 举报

jimmy112595 幼苗

共回答了24个问题采纳率:87.5% 举报

1.计算
(1)(-12a^2b^2c)· (-4分之1abc^2)^2
=(-12a^2b^2c)· (16分之1a^2b^2c^4)
=-3/4a^4b^4c^5
(2)(-a^2)·(-2ab)+3a^2(ab-3分之1b-1)
=2a^3b+3a^3b-a^2b-3a^2
=5a^3b-a^2b-3a^2
(3)8x^2-5x(4y-x)+4x(-4x-2分之5y)
=8x^2-20xy+5x^2-16x^2-10xy
=-30xy-3x^2
(4)(3a^2-2b^2)(a^2+b^2)
=3a^4+3a^2b^2-2a^2b^2-2b^4
=3a^4+a^2b^2-2b^4
2.解方程
1.(2x+5)(x-1)=2(x+4)(x-3)
2x^2-2x+5x-5=2x^2-6x+8x-24
x=-19
3.1.已知x+y=m,
(x+y)^3+(2x+2y)^3+(3x+3y)^3的值
=(x+y)^3+2^3(x+y)^3+3^3(x+y)^3
=(1+8+27)(x+y)^3
=36(x+y)^3
=36m^3

1年前

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