1.(x+x^0)^-1/x^-1+x^0
答:x/(x+1)^2
2.m^(-2)n^(-2)/m^(-2)-n^(-2)
答:1/n^2-m^2
第1,2...知答案唔識計,,thz
指數定律laws of indice
2007-07-31 10:12 am
回答 (3)
2007-07-31 10:50 am
✔ 最佳答案
1.(x + x^0)^(-1) / x^(-1) + x^0
= (x + 1)^(-1) / [(1/x) + 1]
= (x + 1)^(-1) / [(x + 1) / x]
= x / (x + 1)(x + 1)
= x / (x + 1)^2
2.
m^(-2) n^(-2) / m^(-2) - n^(-2)
= [m^(-2) - n^(-2) / m^(-2) n^(-2)]^(-1)
= {[m^(-2) / m^(-2) n^(-2) ] - [n^(-2) / m^(-2) n^(-2)]}^(-1)
= [1/n^(-2) - 1/m^(-2)]^(-1)
= (n^2 - m^2)^(-1)
= 1 / (n^2 - m^2)
2007-07-31 10:56 am
1.(x+x^0)^-1/x^-1+x^0
= [(x + 1)^-1] / [(x^-1) + 1]
= [ 1 / (x + 1)] x {1 / [(x^ -1) - 1] }
= 1 / {(x + 1) [1/(x + 1)]}
= 1 / {(x + 1) [(x + 1) / x ]}
= 1 / {[(x+1)^2] / x}
= x / (x+1)^2
2.m^(-2)n^(-2)/m^(-2)-n^(-2)
= (1/m^2)(1/n^2) / [(1/m^2) - (1/ n^2)]
= (1/m^2 n^2) / [(n^2 - m^2) / (m^2 n^2)]
= (1/m^2 n^2) (m^2 n^2 / n^2 - m^2)
= 1 / n^2 - m^2
= [(x + 1)^-1] / [(x^-1) + 1]
= [ 1 / (x + 1)] x {1 / [(x^ -1) - 1] }
= 1 / {(x + 1) [1/(x + 1)]}
= 1 / {(x + 1) [(x + 1) / x ]}
= 1 / {[(x+1)^2] / x}
= x / (x+1)^2
2.m^(-2)n^(-2)/m^(-2)-n^(-2)
= (1/m^2)(1/n^2) / [(1/m^2) - (1/ n^2)]
= (1/m^2 n^2) / [(n^2 - m^2) / (m^2 n^2)]
= (1/m^2 n^2) (m^2 n^2 / n^2 - m^2)
= 1 / n^2 - m^2
2007-07-31 10:47 am
1.( x + x^0 )^ -1 / x^-1+ x^0
= ( x + 1 )^ -1 / x^ -1 + 1
= ( x + 1 )^ -1 / ( 1/x ) + 1
= ( x + 1 )^ -1 / [ ( 1 + x ) / x ] ∵ 通分母
= x / ( x + 1 ) ² ∵x 當下面的分母時要調去上面
2.m^ (-2) n^ (-2) / m^ (-2) - n^ (-2)
= 1 / m^ (-2) - n^ (-2) [ m ² n ² ] 相乘
= 1 / n ² - m ² ∵m^-2+2 & n^ -2+2
希望幫到你
= ( x + 1 )^ -1 / x^ -1 + 1
= ( x + 1 )^ -1 / ( 1/x ) + 1
= ( x + 1 )^ -1 / [ ( 1 + x ) / x ] ∵ 通分母
= x / ( x + 1 ) ² ∵x 當下面的分母時要調去上面
2.m^ (-2) n^ (-2) / m^ (-2) - n^ (-2)
= 1 / m^ (-2) - n^ (-2) [ m ² n ² ] 相乘
= 1 / n ² - m ² ∵m^-2+2 & n^ -2+2
希望幫到你
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