x/(3^3+4^3)+y/(3^3+6^3)=1 x/(5^3+4^3)+y/(5^3+6^3)=1 求证x+y=432 不得用三角函数!
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x/(3^3+4^3)+y/(3^3+6^3)=1 x/(5^3+4^3)+y/(5^3+6^3)=1 求证x+y=432 不得用三角函数!
x/(3^3+4^3)+y/(3^3+6^3)=1 x/(5^3+4^3)+y/(5^3+6^3)=1 求证x+y=432 不得用三角函数!
x/(3^3+4^3)+y/(3^3+6^3)=1 x/(5^3+4^3)+y/(5^3+6^3)=1 求证x+y=432 不得用三角函数!
【1】去分母,得:(3^3+6^3)x+(3^3+4^3)y=(3^3+4^3)(3^3+6^3),且(5^3+6^3)x+(5^3+4^3)y=(5^3+6^3)(5^3+4^3).两式相减,得:(5^3-3^3)x+(5^3-3^3)y=(4^3+5^3+6^3)5^3-(3^3+4^3+6^3)3^3.【2】(x+y)=[(4^3+5^3+6^3)5^3-(3^3+4^3+6^3)3^3]/(5^3-3^3)=432.
=432
按你写的条件,3^3=5^3.
可能吗?
已知(X/(3^3)+(4^3))+(Y/(3^3)+(6^3))=1,(X/(5^3)+(4^3))+(Y/(5^3)+(6^3))=1.
视(3^3),(5^3)为方程(X/t+(4^3))+(Y/t+(6^3))=1的两根.
化简得(t^2)+((4^3)+6(^3)-X-Y)t+(4^3)•(6^3)-(4^3)Y-6(^3)X=0,
由韦达定理得(3^3)+(5^3)=-((4^3)+6(^3)-X-Y)
∴X+Y=(3^3)+(5^3)+(4^3)+6(^3)
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