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中四數學~變分(~20分~)
(x^2-y^2)α (x^2+y^2),証明y α x
最佳解答:
(x^2-y^2)α (x^2+y^2) (x^2-y^2)=k(x^2+y^2) [k是常數] x^2-y^2=kx^2+ky^2 (1-k)x^2=(k+1)y^2 y^2/x^2=(1-k)/(k+1) (y/x)=√[(1-k)/(k+1)] 令常數c=√[(1-k)/(k+1)] y/x=c y=cx 所以y α x 2007-04-22 02:21:49 補充： 補充:因為k是常數﹐所以c=√[(1-k)/(k+1)]也是常數
其他解答:
(x^2-y^2)a (x^2+y^2),??y a x (x^2-y^2)α (x^2+y^2) (x^2-y^2)=k(x^2+y^2) [k是常數] x^2-y^2=kx^2+ky^2 (1-k)x^2=(k+1)y^2 y^2/x^2=(1-k)/(k+1) (y/x)=√[(1-k)/(k+1)] 令常數c=√[(1-k)/(k+1)] y/x=c y=cx so it was y x a it was not very hard for this quition, so make sure to think deep about it befroe you ask about it next time!!!!
中四數學~變分(~20分~)
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發問:(x^2-y^2)α (x^2+y^2),証明y α x
最佳解答:
(x^2-y^2)α (x^2+y^2) (x^2-y^2)=k(x^2+y^2) [k是常數] x^2-y^2=kx^2+ky^2 (1-k)x^2=(k+1)y^2 y^2/x^2=(1-k)/(k+1) (y/x)=√[(1-k)/(k+1)] 令常數c=√[(1-k)/(k+1)] y/x=c y=cx 所以y α x 2007-04-22 02:21:49 補充： 補充:因為k是常數﹐所以c=√[(1-k)/(k+1)]也是常數
其他解答:
(x^2-y^2)a (x^2+y^2),??y a x (x^2-y^2)α (x^2+y^2) (x^2-y^2)=k(x^2+y^2) [k是常數] x^2-y^2=kx^2+ky^2 (1-k)x^2=(k+1)y^2 y^2/x^2=(1-k)/(k+1) (y/x)=√[(1-k)/(k+1)] 令常數c=√[(1-k)/(k+1)] y/x=c y=cx so it was y x a it was not very hard for this quition, so make sure to think deep about it befroe you ask about it next time!!!!
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