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1
(a+5)(a-5)x(a-5)²/(a+5)²=(a-5)³/(a+5);
2
m-2n=0,m=2n
[1+n/m-m/(m-n)]/[1-n/m-m/(m+n)]
=[1+n/(2n)-2n/(2n-n)]/[1-n/(2n)-2n/(2n+n)]
=[1+1/2-2]/[1-1/2-2/3]
=(-1/2)/(-1/6)
=3
(a+5)(a-5)x(a-5)²/(a+5)²=(a-5)³/(a+5);
2
m-2n=0,m=2n
[1+n/m-m/(m-n)]/[1-n/m-m/(m+n)]
=[1+n/(2n)-2n/(2n-n)]/[1-n/(2n)-2n/(2n+n)]
=[1+1/2-2]/[1-1/2-2/3]
=(-1/2)/(-1/6)
=3
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1、(a²-25)÷[(a²+10a+25)/(25-10a+a²)]
=(a²-25)×(a²-10a+25)/(a²+10a+25)]
=(a-5)(a+5)*(a-5)²/(a+5)²
=(a-5)³/(a+5)
2、[1+n/m-m/(m-n)]÷[1-n/m-m/(m+n)]
=[m(m-n)+n(m-n)-m²]/[m(m-n)]÷{[m(m+n)-n(m+n)-m²]/[m(m+n)]}
=-n²/[m(m-n)]÷[-n²/[m(m+n)]}
=(m+n)/(m-n)
∵m=2n
∴原式=(2n+n)/(2n-n)=3
=(a²-25)×(a²-10a+25)/(a²+10a+25)]
=(a-5)(a+5)*(a-5)²/(a+5)²
=(a-5)³/(a+5)
2、[1+n/m-m/(m-n)]÷[1-n/m-m/(m+n)]
=[m(m-n)+n(m-n)-m²]/[m(m-n)]÷{[m(m+n)-n(m+n)-m²]/[m(m+n)]}
=-n²/[m(m-n)]÷[-n²/[m(m+n)]}
=(m+n)/(m-n)
∵m=2n
∴原式=(2n+n)/(2n-n)=3
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