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已知2x^3=3y^3=4z^3,且1/x+1/y+1/z=1,求开立方(2x^2+3y^2+4z^2)的值
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已知2x^3=3y^3=4z^3,且1/x+1/y+1/z=1,求开立方(2x^2+3y^2+4z^2)的值
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答案和解析
t=2x^3=3y^3=4z^3,x=(t/2)^(1/3),y=(t/3)^(1/3),z=(t/4)^(1/3).
1=1/x+1/y+1/z=(2/t)^(1/3)+(3/t)^(1/3)+(4/t)^(1/3)=[2^(1/3)+3^(1/3)+4^(1/3)]/t^(1/3),
t^(1/3)=2^(1/3)+3^(1/3)+4^(1/3).
2x^2+3y^2+4z^2=t/x + t/y + t/z = t[1/x+1/y+1/z] = t,
[2x^2+3y^2+4z^2]^(1/3)=t^(1/3) = 2^(1/3)+3^(1/3)+4^(1/3)
1=1/x+1/y+1/z=(2/t)^(1/3)+(3/t)^(1/3)+(4/t)^(1/3)=[2^(1/3)+3^(1/3)+4^(1/3)]/t^(1/3),
t^(1/3)=2^(1/3)+3^(1/3)+4^(1/3).
2x^2+3y^2+4z^2=t/x + t/y + t/z = t[1/x+1/y+1/z] = t,
[2x^2+3y^2+4z^2]^(1/3)=t^(1/3) = 2^(1/3)+3^(1/3)+4^(1/3)
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