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x3 1y 1 0 2x 2y 7

P=1+3x^2y-2y,Q=x^3-2x Py=3x^2-2 Qx=3x^2-2 积分与路径无关,选取L1:(1,0)到(0,0)L2:(0,0)到(0,1) ∫L(1+3x2y-2y)dx+(x3-2x)dy =∫L1(1+3x2y-2y)dx+(x3-2x)dy+∫L2(1+3x2y-2y)dx+(x3-2x)dy =∫(0,1)(1)dx+∫(0,1)(0)dy =1

由y=x3-2x+1,得y′=3x2-2.∴y′|x=1=1.∴曲线y=x3-2x+1在点(1,0)处的切线方程为y-0=1×(x-1).即x-y-1=0.故答案为:x-y-1=0.

(1)根据题意得:3-n=1且m-2=2,解得:n=2,m=4;(2)(x3y2+2x2y3)÷12xy2=2x2+4xy,当x=2,y=-12时,原式=2×22+4×2×(-12)=8-4=4.

因为|A|等于特征值的乘积

由题意得,2x+1=0,y-3=0,解得x=-12,y=3,所以,x3+y3=(-12)3+33=-18+27=2678.故答案为:2678.

a^2-4ab+4b^2-2a+4b+1=0 (a-2b)^2-2(a-2b)+1=0 {(a-2b)-1}^2=0 a-2b=1 (a-2b)^2011=1 x^2+y^2-4x-10y+29=0 x^2-4x+4+y^2-10y+25=0 (x-2)^2+(y-5)^2=0 x=2 y=5 x^2y^2+2x^3y^2+x^4y^2 =x^2y^2(x^2+2x+1) =x^2y^2(x+1)^2 =4x25x3^2 =900

∵x2+x-1=0,∴x2+x=1,∴x3+2x2-7=x(x2+x)+x2-7=x+x2-7=1-7=-6.故答案为:-6.

∵x2+x-1=0,∴x2+x=1,∴x3+2x2-7,=x3+x2+x2-7,=x(x2+x)+x2-7,=x+x2-7,=-6.故本题答案为:-6.

解:根据约束条件画出可行域,∵设k=x+2y+3x+1=1+2y+2x+1,整理得(k-1)x-2y+k-3=0,由图得,k>1.设直线l0=(k-1)x-2y+k-3,当直线l0过A(0,3)时l0最大,k也最大为9,当直线l0过B(0,0))时l0最小,k也最小为3.故答案为:[3,9].

满足约束条件x≥0y≥x3x+4y≤12的可行域,如下图中阴影部分所示:∵x+2y+3x+1=2(y+1x+1)+1,表示动点(x,y)与P(-1,-1)点连线斜率的2倍再加1,由图可得当x=0,y=3时,x+2y+3x+1的最大值是9,故选:A

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