用最速下降法求解(x)=({x)_(1)}^2+2(x)_(1)(x)_(2)+2({x)_(2)}^2-4(x)_(1)-3(x)_(2),取(x)=({x)_(1)}^2+2(x)_(1)(x)_(2)+2({x)_(2)}^2-4(x)_(1)-3(x)_(2),迭代两次.

例4 判断二次型-|||-((x)_(1),(x)_(2),(x)_(3))=(x)_(1)^2+2(x)_(2)^2+4(x)_(3)^2+2(x)_(1)(

二次型 ((x)_(1),(x)_(2),(x)_(3))=2({x)_(1)}^2+2({x)_(2)}^2+2({x)_(3)}^2-2(x)_(2)(x)

判定下列二次型的正定性:-|||-(1) =-2({x)_(1)}^2-6({x)_(2)}^2-4({x)_(3)}^2+2(x)_(1)(x)_(2)+2(

21.设二次型((x)_(1),(x)_(2),(x)_(3))=(a)_(1)({x)_(1)}^2+(a)_({x)_(2)}^2+(a-1)({x)_(3

7.已知实二次型f(x1,x2 ,(x)_(3))=9({x)_(1)}^2+2({x)_(2)}^2+({x)_(3)}^2+2lambda (x)_(1)(

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

求曲线=dfrac (3{x)^2-2x+1}({x)^2+2}在点=dfrac (3{x)^2-2x+1}({x)^2+2}处的切线方程和法线方程.求曲线在点

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)-

3.求解线性方程组 ) (x)_(1)+2(x)_(2)-(x)_(3)+2(x)_(4)=1 2(x)_(1)+4(x)_(2)+(x)_(3)+(x)_(

求解下列齐次线性方程组-|||- ) (x)_(1)+2(x)_(2)-(x)_(3)+(x)_(4)=1 2(x)_(1)+4(x)_(2)-2(x)_(3