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$解：原式=[(a+\frac {1}{2}b)(a-\frac {1}{2}b)]2$
$=(a2-\frac {1}{4}b2)2$
$=a^{4}-\frac {1}{2}a2b2+\frac {1}{16}b^{4}$
解：原式???$=[(-m-3)(m-3)]2$???
???$ =(9-m2)2$???
???$ = m^4-18m2+81$???
解：原式???$=[a+(b+c)][a-(b+c)]$???
???$ =a2-(b+c)2$???
???$ =a2-b2-2bc-c2$???
?$ 解：原式=(x+y)2-4z2$?
?$=x2+y2+2xy-4z2$?
?$ 解：原式=[(\frac {1}{2}x-1)(\frac {1}{2}x+1)]2(\frac {1}{4}x2+1)2$?
?$=(\frac {1}{4}x2-1)2(\frac {1}{4}x2+1)2$?
?$=(\frac {1}{16}x^{4}-1)2$?
?$=\frac {1}{256}x^{16}-\frac {1}{8}x^{4}+1$?
解：原式??$=x2-y2-x2-2xy-y2+2y2-2xy$??
??$ =-4xy$??
當(dāng)??$x=1，$????$y=3$??時，
原式??$=-4×1×3=-12$??
?$解：??(1) x+\frac 1x=3，??則??(x+\frac 1x)2=9??$?
?$∵?? (x+\frac 1x)2=x2+\frac 1{x2}+2??$?
?$∴?? x2+\frac 1{x2}=9-=7??$?
?$??(2) (x-\frac 1x)2=x2+\frac 1{x2}-2=7-2=5??$?

$解：原式=[a-(b-c)][a+(b-c)]$
$=a2-(b-c)2$
$=a2-(b2-2bc+c2)$
$=a2-b2+2bc-c2$
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