先化简,在求值(4)(3x²+3yz²-xyz)+2xyz-3yz²+x²z,其中,x=2,y=-1,z=1

Question

先化简,在求值 (4) (3x²+3yz²-xyz)+2xyz-3yz²+x²z,其中,x=2,y= -1,z=1

Answer 1

4(3x²+3yz²-xyz)+2xyz-3yz²+x²z
=12x²+12yz²-12xyz+2xyz-3yz²+x²z
=12x²+12yz²-3yz²-12xyz+2xyz+x²z
=12x²+9yz²-10xyz+x²z
=12x²+x²z+9yz²-10xyz
=x²(12+z)+yz(9z-10x)
=2²(12+1)+(-1)*1*(9*1-10*2)
=4*13-(9-20)
=52-(-11)
=52+11
=63