谁可以给我答案!..=【(根号下K+1)-(根号下K-1)】分之【(根号下K+1)+(根号下K-1)】Y=【(根号下K+1)+(根号下K-1)】分之【(根号下K+1)-(根号下K-1)】K为整数,且X-XY+Y=1,求K的值X=【(根号下K+1)-
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![谁可以给我答案!..=【(根号下K+1)-(根号下K-1)】分之【(根号下K+1)+(根号下K-1)】Y=【(根号下K+1)+(根号下K-1)】分之【(根号下K+1)-(根号下K-1)】K为整数,且X-XY+Y=1,求K的值X=【(根号下K+1)-](/uploads/image/z/5348619-27-9.jpg?t=%E8%B0%81%E5%8F%AF%E4%BB%A5%E7%BB%99%E6%88%91%E7%AD%94%E6%A1%88%21..%3D%E3%80%90%28%E6%A0%B9%E5%8F%B7%E4%B8%8BK%2B1%29-%28%E6%A0%B9%E5%8F%B7%E4%B8%8BK-1%29%E3%80%91%E5%88%86%E4%B9%8B%E3%80%90%EF%BC%88%E6%A0%B9%E5%8F%B7%E4%B8%8BK%2B1%EF%BC%89%2B%EF%BC%88%E6%A0%B9%E5%8F%B7%E4%B8%8BK-1%EF%BC%89%E3%80%91Y%3D%E3%80%90%EF%BC%88%E6%A0%B9%E5%8F%B7%E4%B8%8BK%2B1%EF%BC%89%2B%EF%BC%88%E6%A0%B9%E5%8F%B7%E4%B8%8BK-1%EF%BC%89%E3%80%91%E5%88%86%E4%B9%8B%E3%80%90%EF%BC%88%E6%A0%B9%E5%8F%B7%E4%B8%8BK%2B1%29-%28%E6%A0%B9%E5%8F%B7%E4%B8%8BK-1%29%E3%80%91K%E4%B8%BA%E6%95%B4%E6%95%B0%2C%E4%B8%94X-XY%2BY%3D1%2C%E6%B1%82K%E7%9A%84%E5%80%BCX%3D%E3%80%90%28%E6%A0%B9%E5%8F%B7%E4%B8%8BK%2B1%29-)
谁可以给我答案!..=【(根号下K+1)-(根号下K-1)】分之【(根号下K+1)+(根号下K-1)】Y=【(根号下K+1)+(根号下K-1)】分之【(根号下K+1)-(根号下K-1)】K为整数,且X-XY+Y=1,求K的值X=【(根号下K+1)-
谁可以给我答案!..
=【(根号下K+1)-(根号下K-1)】分之【(根号下K+1)+(根号下K-1)】
Y=【(根号下K+1)+(根号下K-1)】分之【(根号下K+1)-(根号下K-1)】
K为整数,且X-XY+Y=1,求K的值
X=【(根号下K+1)-(根号下K-1)】分之【(根号下K+1)+(根号下K-1)】
Y=【(根号下K+1)+(根号下K-1)】分之【(根号下K+1)-(根号下K-1)】
K为整数,且X-XY+Y=1,求K的值
谁可以给我答案!..=【(根号下K+1)-(根号下K-1)】分之【(根号下K+1)+(根号下K-1)】Y=【(根号下K+1)+(根号下K-1)】分之【(根号下K+1)-(根号下K-1)】K为整数,且X-XY+Y=1,求K的值X=【(根号下K+1)-
X=[√(k+1)+√(k-1)]/[√(k+1)-√(k-1)]
Y=[√(k+1)-√(k-1)]/[√(k+1)+√(k-1)]
显而易见,X·Y=1.因此,X-XY+Y=1可化为:X+Y=2.
代入X和Y的两个表达式得:
[√(k+1)+√(k-1)]/[√(k+1)-√(k-1)]+[√(k+1)-√(k-1)]/[√(k+1)+√(k-1)]=2
左右同乘以[√(k+1)+√(k-1)]·[√(k+1)-√(k-1)],得:
[√(k+1)+√(k-1)]²+[√(k+1)-√(k-1)]²=2[√(k+1)-√(k-1)]·[√(k+1)+√(k-1)]
利用完全平方公式和平方差公式得:(过程较简略,可自己动手写写)
[(k+1)+2√(k²-1)+(k-1)]+[(k+1)-2√(k²-1)+(k-1)]=2[(k+1)-(k-1)]
整理得:4k=4.
得k=1.