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a+1分之a+a+1分之1=？（a+1）²分之a＋（a+1）²分之1（x-1）²分之5－（x-1）²分之5x...
a+1分之a+a+1分之1=？ （a+1）²分之a＋（a+1）²分之1 （x-1）²分之5－（x-1）²分之5x
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=a/(a+1)+1/(a+1)
=(a+1)/(a+1)
=1
=a/(a+1)²+1/(a+1)²
=(a+1)/(a+1)²
=1/(a+1)
（x-1）²分之5－（x-1）²分之5x
=(5-5x)/(x-1)²
=-5(x-1)/(x-1)²
=-5/(x-1)
=(a+1)/(a+1)
=1
=a/(a+1)²+1/(a+1)²
=(a+1)/(a+1)²
=1/(a+1)
（x-1）²分之5－（x-1）²分之5x
=(5-5x)/(x-1)²
=-5(x-1)/(x-1)²
=-5/(x-1)
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a+1分之a+a+1分之1
=（a+1)/(a+1)
=1
（a+1）²分之a＋（a+1）²分之1
=(a+1)/(a+1)²
=1/（a+1）
（x-1）²分之5－（x-1）²分之5x
=(5-5x)/(x-1)²
=-5(x-1)/(x-1)²
=-5/(x-1)
=（a+1)/(a+1)
=1
（a+1）²分之a＋（a+1）²分之1
=(a+1)/(a+1)²
=1/（a+1）
（x-1）²分之5－（x-1）²分之5x
=(5-5x)/(x-1)²
=-5(x-1)/(x-1)²
=-5/(x-1)
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a/(a+1)+1/(a+1)
=(a+1)/(a+1)=1
a/(a+1)²+1/(a+1)²
=(a+1)/(a+1)²
=1/(a+1)
5/(x-1)²-5x/(x-1)²
=5(1-x)/(x-1)
=-5(x-1)(x-1)
=-5/(x-1)
=5/(1-x)
=(a+1)/(a+1)=1
a/(a+1)²+1/(a+1)²
=(a+1)/(a+1)²
=1/(a+1)
5/(x-1)²-5x/(x-1)²
=5(1-x)/(x-1)
=-5(x-1)(x-1)
=-5/(x-1)
=5/(1-x)
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