若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/30 11:19:44
若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
答:
x^2-3x+1=0
y^2-3y+1=0
所以:x和y是方程a^2-3a+1=0的两个根
根据韦达定理:
x+y=3
xy=1
所以:
(x+y)^2=9
x^2+2xy+y^2=9
所以:
x^2+y^2=7
所以:
(x^2+y^2)^2=49
x^4+2(xy)^2+y^4=49
x^4+y^4+2=49
解得:
x^4+y^4=47
若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值
若x,y满足|2x-y-3|+|3x+2y+1|=0,则x= y=
若x,y满足|x-y+1|+(x+y+3)^2=0则x^2-y^2=
若(2x-y-1)2+|x-2y-3|=0,求代数式(2x+y)(2x-y)-(x+2y)(x-2y-1)的值若(2x-y-1)²+|x-2y-3|=0,求代数式(2x+y)(2x-y)-(x+2y)(x-2y-1)的值
若变量x,y满足约束条件{x>=-1;y>=x;3x+2y
若x=-1/4,能否确定代数式(2x-y)(2x+y)+(2x-y)(y-4x)+2y(y-3x)的值?
若x+y=4,则多项式1/2(x+y)+4(x-y)-3(x-y)-3/2(x+y)-x+y值是()
(3x-y)^2+(3x+y)(3x-y),x=1,y=-2
二元一次方程 :2(x+y)-(x-y)=3 (x+y)-2(x-y)=1
(x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5)
2(x+y)-3(x-y)=1 6(x+y)+(x-y)=51
{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1
变量x,y满足约束条件,x+y>=3,x-y>=-1,2x-y
(x-y)^2+(x+y)(x-y) 其中 X =3 Y=-1
若2x-3y+4=0则x(x*x-1)+x(5-x*x)-6y+7
{4x-3y-10=0 3x-2y=0,{3x-4(x-y)=2 2x-3y=1,{2(x+y)-(x-y)=3 (x+y)-2(x-y)=1
已知2x-y=0,求分式【1+(3*y*y*y/x*x*x-y*y*y)】/【1+(2y/x-y)】的值注:x*x*x-y*y*y=(x-y)*(x*x+xy+y*y)
若y=x(x-1)(x-2)(x-3),则y′(0)=