x²-4x+2=0
解方程 x(x+1)(x²-2x-4)=0 x³-2x+1=0 x的4次方+8x³+14x
2x²+4x+x=0,求x²+2x的算术平方根
设f (x)={2x+1(x≥0),x²+4(x
x^4+x^3+x^2+x+1=0,x^2006+x^2005+x^2004+x^2003+x^2002
因为x²-x+2=(x-1/2)²+7/4>0所以x(x²-x+2)
x²-4x+2=0
已知f(x)是偶函数,x≥0时,f(x)=-2x²=4x,求x
x²+3x-2=0,x²-6x-6=0,3x²-4x-1=0,3x²+10x+3
3x²-(x-2)=0 (2x-1)(x+3)=4 x²-3x-4=0 x²-3x+18=
0.3x²+x=0.8 x²+3=2√3x x²+2x-2=0 3x²+4x-7
(x-2)²=(2x-3)² x²-4x=0 3x(x+1)=3x+3 x²-2
2x(x-1)=0 x²+4=4x (x-3)(x+1)=0 x²-3x+1=0