y=(x^2-4x-5)/(x^2-3x-4)的值域是答案是y≠1且y≠6/5怎么算的?
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/08 19:31:11
y=(x^2-4x-5)/(x^2-3x-4)的值域是答案是y≠1且y≠6/5怎么算的?
y=(x^2-4x-5)/(x^2-3x-4)的值域是
答案是y≠1且y≠6/5
怎么算的?
y=(x^2-4x-5)/(x^2-3x-4)的值域是答案是y≠1且y≠6/5怎么算的?
x^2-3x-4≠0,(x-4)(x+1)≠0,x≠4且x≠-1
y = (x-5)(x+1)/(x-4)(x+1)=(x-5)/(x-4) (用y来表示x,两边乘x-4)
yx-4y=x-5,(y-1)x=4y-5,x=(4y-5)/(y-1),(y≠1)
所以(4y-5)/(y-1)≠4,恒成立,(4y-5)/(y-1)≠-1,5y≠6,y≠6/5
所以y≠1且y≠6/5
若2x-3y+4=0则x(x*x-1)+x(5-x*x)-6y+7
3(x+y)+2(x-y)=36 4(x+y)-5(x-y)=2
4(x-y)-3(x-y)=9 5(x-y)+2(x-y)=11
3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27
先化简再求值(x-y)(x-2y)+(x-2y)(x-3y)-2(x-3y)(x-4y) x=4 y=5
xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1
(x-y) (x-2y)+(x+2y) (x-3y)-2(x-3y) (x-4y)=
已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y)
3X=5Y (X+Y)*3X=(2Y+X)*4+2*2
因式分解(x-2y)(x-3y)+(x-4y)(x-5y)+(x-3y)(x-5y)+(x-2y)(x-4y)
*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X ||
3^x+2x=y,x
化简x-2y+3x-4x+5x-...+2001x-2002x,急.
化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1
求值域 1,y=(x+2)/(x*x+3x+6) 2,y=3x/(x*x+4)
已知7x+5y/2y-x=3/2,且x≠0,求X/x-4y
{x+y=1 ,xy=-6{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0
1.(x^3-x^2-4x+1)/(x^2-3x+2)-(x^3-2x^2-9x+21)/(x^2-5x+6)+(x^2-3x+8)/(x^2-4x+3)2.{(x^2)/(x-y)}*{y/(x+y)}-{(x^4y)/(x^4-y^4)}÷{(x^2/x^2+y^2)^2}