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若X+Y=3,XY=1,X的5次方+Y的5次方得几
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若X+Y=3,XY=1,X的5次方+Y的5次方得几
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答案和解析
x+y=3
(x+y)²=9
x²+y²+2xy=9
x²+y²=7
x^5+y^5
=(x+y)(x^4-x^3*y+x^2*y^2-x*y^3+y^4)
=3[x^4+y^4-xy(x^2+y^2)+(xy)^2]
=3[(x²+y²)²-2(xy)²-xy(x²+y²)+(xy)² ]
= 3[7²-2*1²-1*7+1]
=3*(49-2-7+1)
=3*41
=123
=19/4
(x+y)²=9
x²+y²+2xy=9
x²+y²=7
x^5+y^5
=(x+y)(x^4-x^3*y+x^2*y^2-x*y^3+y^4)
=3[x^4+y^4-xy(x^2+y^2)+(xy)^2]
=3[(x²+y²)²-2(xy)²-xy(x²+y²)+(xy)² ]
= 3[7²-2*1²-1*7+1]
=3*(49-2-7+1)
=3*41
=123
=19/4
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