Question

数学CASE若分式方程x^2-1分之k-1减去x^2-x分之1=x^2+x分之k-5有增根x=-1,求K的值当x=200

Answer 1

1.

通分有：

[x(k-1)-(x+1)-(k-5)(x-1)]/[x(x-1)(x+1)]=0

→x=(6-k)/3

∵有增根-1

∴有(6-k)/3=-1

k=9

2.根据立方差公式有

(x+3)(x²-3x+9)-(x-2)(x²+2x+4)=(x³+3³)-(x³-2³)=3³-2³

=27-8=19

3.

x≥y,所以|x-y|=x-y

x+y-(x-y)=2y

4.{5a^2+(6a)-7}+{(-4)a^2-4a+(8)}=a^2+2a+1

5.(x-a)(x-b)=x^2+AX+B,则A=__B=___

A=-(a+b),B=ab

6.设n为正整数,且x^2n=2,计算(3x^3n)^2-4(x^2)^2n的值

(3x^3n)^2-4(x^2)^2n=3^2 * x^6n-4*x^4n

=9*(x^2n)^3-4*(x^2n)^2=9*2^3-4*2^2=72-16=56