初二一次函数难题急需

题

1.已知(-3x^4y^3)^3÷(-3/2x^ny²)=-mx^8y^7,求n，m的值

2.已知3x-2y=5,求[6x^4+(x^3-y)²-(x^3+y)²]÷(-2x)^3的值

3.因式分解计算
(1)31×3.14+27×3.14+42×3.14
(2)27×2.78+73×2.78
(3)当x=2/5,y=7/20,z=1/4时，求xyz²+xy²z+x²yz的值
(4)先分解因式，再求值。2xy²-2x²y,其中y-x=3,xy=1/6

答

1. (-3x^4y^3)^3÷(-3/2x^ny²)=18x^(12-n)y^7=-mx^8y^7

则有：-m=18 及12-n=8

所以：m=-18 n=4

2. [6x^4+(x^3-y)²-(x^3+y)²]÷(-2x)^3=[6x^4+(x^3-y+X^3+y)(x^3-y-X^3-y)]÷(-2x)^3

=(6x^4-4x^3y)÷(-2x)^3

=-(3x-2y)

当3x-2y=5时，[6x^4+(x^3-y)²-(x^3+y)²]÷(-2x)^3=-5

3. (1) 原式=3.14×(31+27+42)=314

(2)原式=2.78×(27+73)=278

(3)xyz=7/200 x+y+z=1

xyz²+xy²z+x²yz=xyz(x+y+z)=7/200

(4)2xy²-2x²y=2xy(y-x)=2×1/6×3=1