已知m=(√2-1)负一次方+(√2+1)负一次方,那么(m-1)负一次方+(n+1)负一次方等于多少?
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已知m=(√2-1)负一次方+(√2+1)负一次方,那么(m-1)负一次方+(n+1)负一次方等于多少?
答:
m=(√2-1)^(-1)+(√2+1)^(-1)
=1/(√2-1)+1/(√2+1)
=(√2+1)/[(√2+1)(√2-1)]+(√2-1)/[(√2-1)(√2+1)]
=(√2+1)/(2-1)+(√2-1)/(2-1)
=√2+1+√2-1
=2√2
(m-1)^(-1)+(m+1)^(-1)
=1/(m-1)+1/(m+1)
=(m+1+m-1)/(m²-1)
=2m/(m²-1)
=2*2√2/(8-1)
=4√2/7
m=(√2-1)^(-1)+(√2+1)^(-1)
=1/(√2-1)+1/(√2+1)
=(√2+1)/[(√2+1)(√2-1)]+(√2-1)/[(√2-1)(√2+1)]
=(√2+1)/(2-1)+(√2-1)/(2-1)
=√2+1+√2-1
=2√2
(m-1)^(-1)+(m+1)^(-1)
=1/(m-1)+1/(m+1)
=(m+1+m-1)/(m²-1)
=2m/(m²-1)
=2*2√2/(8-1)
=4√2/7