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電子課本網(wǎng) 第35頁(yè)

第35頁(yè)

信息發(fā)布者:
C
D
11
4049
16
解:?$ $?因?yàn)?$(x - 2y)^2=(x + 2y)^2-8xy,$?
又?$x + 2y = 2,$??$xy=-1,$?
所以原式?$=2^2-8×(-1)=4 + 8 = 12。$?
解:
?$ \begin {aligned}&x^4+16y^4\\=&(x^2 + 4y^2)^2-8(\mathrm {xy})^2\\=&[(x + 2y)^2-4xy]^2-8(\mathrm {xy})^2\end {aligned}$?
?$ $?因?yàn)?$x + 2y = 2,$??$xy=-1,$?
所以原式?$=[2^2-4×(-1)]^2-8×(-1)^2$?
?$=8^2-8 = 56$?
解:因?yàn)?$m + n = 8,$??$mn = 15,$?
所以?$(m - n)^2=(m + n)^2-4mn=64 - 60 = 4,$?
則?$m - n = 2,$?
所以?$(m + n)(m - n)=\mathrm {m^2} - n^2 = 16,$?
所以?$S_{1}-S_{2}=\mathrm {m^2}-S_{空白}-(n^2-S_{空白})$?
?$=\mathrm {m^2} - n^2 = 16$?