9x^4+12x^2y^2+4y^4 9-12(x+y)+4(x+y)^2 x^2+2x(y+z)+(y+z)^2 36
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9x^4+12x^2y^2+4y^4 9-12(x+y)+4(x+y)^2 x^2+2x(y+z)+(y+z)^2 36^2+72*14+14^2
9x^4+12x²y²+4y^4
=(3x²)²+12x²y²+(2y²)²
=(3x²+2y²)²
9-12(x+y)+4(x+y)²
=3²-12(x+y)+[2(x+y)]²
=(2x+2y-3)²
x²+2x(y+z)+(y+z)²
=(x+y+z)²
36²+72*14+14²
=36²+2*36*14+14²
=(36+14)²
=50²
=2500
=(3x²)²+12x²y²+(2y²)²
=(3x²+2y²)²
9-12(x+y)+4(x+y)²
=3²-12(x+y)+[2(x+y)]²
=(2x+2y-3)²
x²+2x(y+z)+(y+z)²
=(x+y+z)²
36²+72*14+14²
=36²+2*36*14+14²
=(36+14)²
=50²
=2500
{(z+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y
设x、y、z为整数,证明:x^4*(y-z)+y^4*(z-x)+z^4*(x-y)/(y+z)^2+(z+x)^2+(
x=y/z=z/3,x+y+z =12,求2x+3y+4z是多少,
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai
求解2x≥x+y+z 4y≥x+y+Z 8z≥x+y+z x、y、z均大于0
①x+y+z=6 3x-y+2z=12 x-y-3z=-4 ②x+y-z=2 4x-2y+3y+8=0 x+3y-2z-
1.x+y=16,y+z=12,z+x=102.3x-y+z=4,2x+3y-z=12,x+y+z=63.x+y+z=6
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x
三元一次方程组x-2y=-9 3x-y+z=4 y=2x-7 y-z=3 2x+3y-z=12 5x+3y+2z=2 2
解方程组{3x+y-z=4,2x-y+3z=12,x+y+z=6}