以下哪项金融产品是以“1X3”、“3X6”、“3X9”的方式来描述的（）。

求函数y=x3-3x2-9x+1的极值.正确答案:由于y=x3-3x2-9x+1的定义域为(-∞,+∞),y’=3x2-6x-9,令y’=0,得驻点x1=-1,

线性方程组 ) (x)_(1)+2(x)_(2)-2(x)_(3)=1 2(x)_(1)+4(x)_(2)-4(x)_(3)=2 3(x)_(1)+6(x)_

例4 讨论线性方程组-|||- ) (x)_(1)+(x)_(2)+2(x)_(3)+3(x)_(4)=1 (x)_(1)+3(x)_(2)+6(x)_(3)

函数=dfrac (1)({3)^x-9}的定义域是（ ）=dfrac (1)({3)^x-9}=dfrac (1)({3)^x-9}=dfrac (1)(

五、给出线性规划问题:(20)Max z=2x1+4x2+x3+x4 x1+3x2+x4≤8 2x1+x2≤6 x2+x3+x4≤6 x1+x2+x3≤9 xj

3.已知方程组 ) (x)_(1)+(x)_(2)+2(x)_(3)=a 3(x)_(1)-(x)_(2)-6(x)_(3)=a+2 (x)_(1)+4(x

((x)^3+1)=(x)^6+3(x)^3+2 ，则 f ( x ) =，则f(x)=

用列主元消去法解方程组 ) 3(x)_(1)-(x)_(2)+4(x)_(3)=1 -(x)_(1)+2(x)_(2)-9(x)_(3)=0 -4(x)_(1

设dfrac(x)({x)^2 -3x+1} =1，求dfrac({x)^3}({x)^6-27(x)^3+1}的值.设$\dfrac{x}{{x}^{2} -

) (x)_(1)+4(x)_(2)+2(x)_(3)geqslant 8 3(x)_(1)+2(x)_(2)geqslant 6 (x)_(1),(x)_(2