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第29页

信息发布者：

$ =(4x^2-1)(4x^2+1)$

$=16x^4-1$

$ =[(2m+3n)(3n-2m)]^2$

$=(9n^2-4\ \mathrm {m^2})^2$

$=81n^4-72\ \mathrm {m^2}n^2+16m^4$

$ =4x^2-y^2-4x^2+4xy-y^2$

$=4xy-2y^2$

$ =(x-y)^2-1^2$

$=x^2-2xy+y^2-1$

解：$Q=(\mathrm {m^2}+1-m)(\mathrm {m^2}+1+m)=(\mathrm {m^2}+1)^2-\mathrm {m^2}$

$=m^4+2\ \mathrm {m^2}+1-\mathrm {m^2}=m^4+\mathrm {m^2}+1$

$P=[(m+1)(m-1)]^2=[\mathrm {m^2}-1]^2=m^4-2\ \mathrm {m^2}+1$

∵$m≠0$

∴$\mathrm {m^2}＞0$

∴$Q-P=m^4+\mathrm {m^2}+1-m^4+2\ \mathrm {m^2}-1=3\ \mathrm {m^2}＞0$

∴$Q＞P$