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多项式3x 2 y%2xy+2y%1是______次______项式,次数...

=-x^3y-5x^2y-3xy^2+2y-1

∵多项式2x2+3xy-2y2-x+8y-6可以分解为(x+2y+m)(2x-y+n)的形式,∴(x+2y+m)(2x-y+n)=2x2+3xy-2y2+(2m+n)x+(2n-m)y+mn=2x2+3xy-2y2-x+8y-6,∴2m+n=-1,2n-m=8,mn=-6,解得m=-2,n=3,∴m3+1n2?1=?8+19?1=-78,故答案为:-78.

∵x+y=-1, ∴x4+5x3y+x2y+8x2y2+xy2+5xy3+y4, =(x4+2x2y2+y4)+5xy(x2+y2)+xy(x+y)+6x2y2, =(x2+y2)2+5xy[(x+y)2-2xy]+xy(x+y)+6x2y2, =[(x+y)2-2xy]2+5xy(1-2xy)-xy+6x2y2, =(1-2xy)2+5xy-10x2y2-xy+6x2y2, =1-4xy+4x2y2+...

B 根据已知得出关于x、y的方程组,求出方程组的解,把x、y的值代入求出即可.解:∵(x﹣y+1) 2 +|2x+y﹣7|=0,∴x﹣y+1=0,2x+y﹣7=0,即 ,①+②得:3x﹣6=0,∴x=2,把x=2代入①得:2﹣y+1=0,∴y=3,∴x 2 ﹣3xy+2y 2 ,=(x﹣y)(x﹣2y),=(2...

x+y=5 xy=4 由此可得 xy2-2x-2y+1 =4*2-2(x+y)+1 =8-2*5+1 =8-10+1=-1

设切点为P(x0,y0,z0),故曲面在切点处的切平面的法向量为n={2x0,2y0,?1}又由于n∥(2,2,1),且切点P在曲面上∴2x02=2y02=?11x02+y02+z0=1解得:x0=y0=-1,z0=-1∴点P处的切平面方程为2(x+1)+2(y+1)+(z+1)=0即2x+2y+z+5=0

不详

如图

(1)原式=-5an-an+7an-3an=-2an;(2)原式=(1+2)3x2y-(3+1)4xy2=3x2y-4xy2;(3)原式=(14-12)a2b-(0.4-25)ab2=?14a2b;(4)原式=3a-2c+6a-c+b+c+a+8b-6=(3+1+6)a+(1+8)b+(-2+1-1)c-6=10a+9b-2c-6.

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