2y^4+3y^3-16y^2+3y+2=0

来源：百度知道编辑：UC知道时间：2024/05/31 17:57:26

解方程

2y^4+3y^3-16y^2+3y+2=0
2y^4 - 4y^3 + 7y^3 - 14y^2 - 2y^2 + 4y - y +2 = 0
2y^3(y-2) + 7y^2(y-2) -2y(y-2) - (y-2) = 0
(y-2)(2y^3 + 7y^2 - 2y -1) = 0
(y-2)(2y^3 - y^2 + 8y^2 - 4y + 2y -1) = 0
(y-2)[y^2(2y-1) + 4y(2y-1) + (2y -1)] = 0
(y-2)(2y-1) (y^2 + 4y + 1) = 0
(y-2)(2y-1)[(y+2)^2 -3] = 0
(y-2)(2y-1)(y + 2 -√3)(y +2 +√3) = 0
所以
y1 =2
y2 = 1/2
y3 = -2 + √3
y4 = -2 - √3

y^5-3y^4+4y^3+2y^3÷y+4= 3(x+y)(x-y)+4(x-y)^2=? (2x+y)(2x-y)-(3x-2y)(x+y)-y(2x-y) 已知-3<y<2.化简:|y-2|+|y+3|-|3y+9|-|2y-4|. (x-2y)^2+(x-y)(x-2y)-2(x-3y)(x-y) x^2+3(x+y)+3-y^2+(x-y) (X-3y)(x+3y)-(x-3y)^2 3(2y+1)=2(1+y) y=? 3X(X-Y)+2Y(X-Y) 4-y×y×(y-3)=0 y=?