17. 计算:
(1)$(x-\frac{1}{2})(x^{2}+\frac{1}{4})(x+\frac{1}{2})$; (2)$(x^{n}+y^{n})(x^{2n}+y^{2n})(x^{n}-y^{n})$.
(1)$(x-\frac{1}{2})(x^{2}+\frac{1}{4})(x+\frac{1}{2})$; (2)$(x^{n}+y^{n})(x^{2n}+y^{2n})(x^{n}-y^{n})$.
答案:
(1)解:原式$=(x-\frac{1}{2})(x+\frac{1}{2})(x^{2}+\frac{1}{4})$
$=(x^{2}-\frac{1}{4})(x^{2}+\frac{1}{4})$
$=x^{4}-\frac{1}{16}$
(2)解:原式$=(x^{n}+y^{n})(x^{n}-y^{n})(x^{2n}+y^{2n})$
$=(x^{2n}-y^{2n})(x^{2n}+y^{2n})$
$=x^{4n}-y^{4n}$
(1)解:原式$=(x-\frac{1}{2})(x+\frac{1}{2})(x^{2}+\frac{1}{4})$
$=(x^{2}-\frac{1}{4})(x^{2}+\frac{1}{4})$
$=x^{4}-\frac{1}{16}$
(2)解:原式$=(x^{n}+y^{n})(x^{n}-y^{n})(x^{2n}+y^{2n})$
$=(x^{2n}-y^{2n})(x^{2n}+y^{2n})$
$=x^{4n}-y^{4n}$
思维与拓展 16
运用平方差公式计算:
$100^{2}-99^{2}+98^{2}-97^{2}+…+2^{2}-1^{2}$.
运用平方差公式计算:
$100^{2}-99^{2}+98^{2}-97^{2}+…+2^{2}-1^{2}$.
答案:
解:原式=(100²-99²)+(98²-97²)+…+(2²-1²)
=(100-99)(100+99)+(98-97)(98+97)+…+(2-1)(2+1)
=1×(100+99)+1×(98+97)+…+1×(2+1)
=(100+99)+(98+97)+…+(2+1)
=(100+1)+(99+2)+…+(51+50)
=101×50
=5050
=(100-99)(100+99)+(98-97)(98+97)+…+(2-1)(2+1)
=1×(100+99)+1×(98+97)+…+1×(2+1)
=(100+99)+(98+97)+…+(2+1)
=(100+1)+(99+2)+…+(51+50)
=101×50
=5050
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