由|2y-1/y-2|≤1怎么推出3y2+8y-3

由|2y-1/y-2|≤1怎么推出3y2+8y-3≤0

3个回答

3^y=x^2+1分之mx^2+8x+n 可推出（3^y-m)x^2-8x=3^y=0
2y^2+3y-2=0是怎么分解成(y+2)(2y-1)=0,
微分方程(1)(y'')^2+5y'-y+x=0;(2)y''+5y'+4y^2-8x=0;(3)(3x+2y)dx+(
3x-2y=-1(1)2x+3y=8(2)x/2-y/3=1/3(1)3(x-1)=y+1(2)x-3y=2(1)y:x
解方程,3y^3+(y+3)(6-2y)+3y(y+2)(2-y)-2(y^2+1)=0
解方程：1/3(3y- 10-7y/2)-1/2(2y-2y+2/3)=y/2-1 0.4y+0.9/0.5-0.03+
(2x^3y^3-2x^3y+1/8^4+y)/(-1/2x^3y)
1/2(y+1)+1/3(y+2)+1/4(y+3)-3=0
3(y-2)-(1-y)=8(y-1)
x+2/3-2y=-8 0.3x-0.1y=-1解方程