已知a+b=7,ab=11,求a^3+b^3和a^3-b^3
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已知a+b=7,ab=11,求a^3+b^3和a^3-b^3
已知a+b=7,ab=11,求a^3+b^3和a^3-b^3
已知a+b=7,ab=11,求a^3+b^3和a^3-b^3
a³+b³=﹙a+b﹚﹙a²-ab+b²
=﹙a+b﹚[﹙a+b﹚²-3ab]
=7×﹙7²-3×11﹚
=102
∵﹙a+b﹚²=7²
∴ ﹙a-b﹚²+4ab=49
﹙a-b﹚²=5
a-b=±√5
a³-b³=﹙a-b﹚﹙a²+ab+b²﹚
=﹙a-b﹚[﹙a+b﹚²-ab]
=±√5﹙7²-11﹚
=±38√5
a^3+b^3
=(a+b)(a^2-ab+b^2)
=(a+b)(a^2+2ab+b^2-3ab)
=(a+b)[(a+b)^2-3ab]
=7*(7^2-3*11)
=112
a^3-b^3
=(a-b)(a^2+ab+b^2)
=(a-b)(a^2+2ab+b^2-ab)
=(a-b)[(a+b)^2-ab]
=(a-b)*(7^2-11)
=38(a-b)
(a-b)^2=(a+b)^2-4ab=49-44=5,
=> a-b=±√5,
=> a^3-b^3=±38√5.
a³+b³=(a+b)(a²-ab+b²)=7((a+b)²-3ab)=7(49-33)=112
a³-b³=﹙a-b﹚﹙a²+ab+b²﹚=(√﹙a-b﹚²)﹙﹙a+b﹚²-ab﹚=(√(a²-2ab+b²))﹙a+b﹚²-ab﹚=(√(a²...
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a³+b³=(a+b)(a²-ab+b²)=7((a+b)²-3ab)=7(49-33)=112
a³-b³=﹙a-b﹚﹙a²+ab+b²﹚=(√﹙a-b﹚²)﹙﹙a+b﹚²-ab﹚=(√(a²-2ab+b²))﹙a+b﹚²-ab﹚=(√(a²-4ab+b²+2ab))﹙﹙a+b﹚²-ab﹚=√27乘﹙﹙a+b﹚²-ab﹚=38√27=114√3
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