已知实数x,y(x≠y),满足方程x^2-7x+1=0,y^2-7y+1=0,那么y/x+x/y的值为多少?
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![已知实数x,y(x≠y),满足方程x^2-7x+1=0,y^2-7y+1=0,那么y/x+x/y的值为多少?](/uploads/image/z/5927194-10-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0x%2Cy%EF%BC%88x%E2%89%A0y%29%2C%E6%BB%A1%E8%B6%B3%E6%96%B9%E7%A8%8Bx%5E2-7x%2B1%3D0%2Cy%5E2-7y%2B1%3D0%2C%E9%82%A3%E4%B9%88y%2Fx%2Bx%2Fy%E7%9A%84%E5%80%BC%E4%B8%BA%E5%A4%9A%E5%B0%91%3F)
已知实数x,y(x≠y),满足方程x^2-7x+1=0,y^2-7y+1=0,那么y/x+x/y的值为多少?
已知实数x,y(x≠y),满足方程x^2-7x+1=0,y^2-7y+1=0,那么y/x+x/y的值为多少?
已知实数x,y(x≠y),满足方程x^2-7x+1=0,y^2-7y+1=0,那么y/x+x/y的值为多少?
由题意x,y是方程x^2-7x+1=0两个根
由违达定理得x+y=7 xy=1
因此y/x+x/y=(x^2+y^2)/(xy)=[(x+y)^2-2xy]/(xy)=(49-2)/1=47
明显x,y是方程x^2-7x+1=0两个解由解的关系有
x+y=7 xy=1求就行
x^2-7x+1=0,
y^2-7y+1=0
两式相减得
x^2-y^2-7(x-y)=0
(x+y-7)(x-y)=0
x+y=7,x-y=0(舍去,(x≠y))
(x+y)^2=49=x^2+y^2+2xy
y/x+x/y
=(y^2+x^2)/(xy)
=(49-2xy)/(xy)
=49/(xy)-2
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