解1题:
原式=(1/4)m³n-(1/4)m²n²+(1/16)mn³
=(1/4)mn[m²-mn+(1/4)n²]
=(1/4)mn[m-(1/2)n]²
2题:
原式=(x²-2x)²+2x(x-2)+1
=(x²-2x)²+2(x²-2x)+1
=[(x²-2x)+1]²
=(x²-2x+1)²
=[(x-1)²]²
=(x-1)的4次方
解1题:
原式=(1/4)m³n-(1/4)m²n²+(1/16)mn³
=(1/4)mn[m²-mn+(1/4)n²]
=(1/4)mn[m-(1/2)n]²
2题:
原式=(x²-2x)²+2x(x-2)+1
=(x²-2x)²+2(x²-2x)+1
=[(x²-2x)+1]²
=(x²-2x+1)²
=[(x-1)²]²
=(x-1)的4次方