由曲线=(x)^2 =(x)^3 ,x=1 x=2所围成的封闭图形面积S表达式是( )。=(x)^2 =(x)^3 ,x=1 x=2=(x)^2 =(x)^3 ,x=1 x=2=(x)^2 =(x)^3 ,x=1 x=2=(x)^2 =(x)^3 ,x=1 x=2

解方程组： ) (x)_(1)-(x)_(2)-(x)_(3)=2 2(x)_(1)-(x)_(2)-3(x)_(3)=1 3(x)_(1)+2(x)_(2)

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

给定线性方程组 ) (x)_(1)+(x)_(2)+(x)_(3)=a-3 (x)_(1)+a(x)_(2)+(x)_(3)=-2 (x)_(1)+(x)_(

用克莱姆法则求解方程组 ) (x)_(1)-(x)_(2)-(x)_(3)=-1 -2(x)_(1)+2(x)_(2)+(x)_(3)=1 2(x)_(1)-

方程组 ) (x)_(1)+(x)_(2)+(x)_(3)=1 (x)_(1)+2(x)_(2)-(x)_(3)=2 (x)_(1)+k(x)_(2)+(x)

2.解方程组 ) (x)_(1)+(x)_(2)+(x)_(3) (x)_(1)+(x)_(2)-(x)_(3)-(x)_(4)=1 5(x)_(1)+5(

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

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)_

解下列线性方程组 ) (x)_(1)+2(x)_(2)+3(x)_(3)=1 2(x)_(1)+2(x)_(2)+5(x)_(3)=2 3(x)_(1)+5(

13.-|||-多项式 ((x)_(1),(x)_(2),(x)_(3))=(({x)_(1)+(x)_(2))}^2+(({x)_(1)+(x)_(3))}^