x 2y=10,ax by=1与方程bx ay=6,2x-y=5的解相同
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(x+y)(xy)=x^2y+xy^2=-8原式=-7
(3x2y-2xy2)-(xy2-2x2y)=3x2y-2xy2-xy2+2x2y=5x2y-3xy2当x=-1,y=2时,原式=5×(-1)2×2-3×(-1)×22=10+12=22.
x=-1,y=3代入②得:-b+3a=1x=3,y=2代入①得:6ab=16解得:(怀疑题目有问题,因为a、b为无理数)求出a、b之后,再代入,求原来的解.
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
原式=4x2y-6xy+3(4xy-2)+x2y+1=5x2y+6xy-5当x=2,y=-12时,原式=5×4×(-12)+6×2×(-12)-5=-21.
化简得:9-12Y^2+6Y+4+12Y^2+4Y-10-10Y+X-Y+1=X-Y+4带入X、Y值得:=3
A+B+C=(x3+3x2y-5xy2+6y3-1)+(y3+2xy2+x2y-2x3+2)+(x3-4x2y+3xy2-7y3+1)=(1+1-2)x3+(3+1-4)x2y+(-5+2+3)xy2
(X+Y)2=1402X2Y*3=14400(X+Y)2=140→X+Y=70→Y=70-X①2X2Y*3=14400→XY=1200②把①代人②得:X(70-X)=1200X²-70X+1
原式=y(x2+2x+1)=y(x+1)2,故答案为:y(x+1)2.
原式=(x^4-2x²y²+y^4)+6xy(x²+2xy+y²)-2xy(x+y)=(x²-y²)²+6xy(x+y)²
原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y,∵(x-2)2+|y+1|=0,∴x-2=0,y+1=0,即x=2,y=-1,则原式=16+16=32.
(x+2)²+|y-1|=0平方数与绝对值都是非负数两个非负数的和为0,那么这两个数都是0x+2=0y-1=0解得:x=-2,y=1x³+3x²y+3xy²+y
原式=2x2y-2xy2-[-3x2y2+3x2y+3x2y2-3xy2]=2x2y-2xy2+3x2y2-3x2y-3x2y2+3xy2=2x2y-3x2y-2xy2+3xy2+3x2y2-3x2y
代入x=-1,y=1,2x^y-(5xy^-3x^y)-x^=2*(-1)^*1-{5*(-1)*1^-3*(-1)^*1}-(-1)^=2-(-5-3)-1=9备注:2^表示2的平方
原式=2x2y+2xy-3x2y+3xy-4x2y=-5x2y+5xy,当x=-1,y=1时,原式=-5×(-1)2×1+5×(-1)×1=-5-5=-10.
解方程组4x+y=53x-2y=1得,x=1,y=1代入另外两个方程a+b=3(这个方程应该是ax+by=3吧)a-b=1解得z=2,b=1
3,8,15,24的差分别是5,7,94,6,8,10差都是25,10,17,26分别是3,8,15,24加2第五个式子应该是35方+12方=37方