数学化简(√6+4*√3+3*√2)/【(√6+√3)(√3+√2)】[(n+1)√n--n*√(n+1)]/√(n+1)-√n
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数学化简(√6+4*√3+3*√2)/【(√6+√3)(√3+√2)】[(n+1)√n--n*√(n+1)]/√(n+1)-√n
数学化简(√6+4*√3+3*√2)/【(√6+√3)(√3+√2)】
[(n+1)√n--n*√(n+1)]/√(n+1)-√n
数学化简(√6+4*√3+3*√2)/【(√6+√3)(√3+√2)】[(n+1)√n--n*√(n+1)]/√(n+1)-√n
(√6+4*√3+3*√2)/【(√6+√3)(√3+√2)】
=【(√6+√3)+3(√3+√2)】/【(√6+√3)(√3+√2)】
=(√6+√3)/【(√6+√3)(√3+√2)】+3(√3+√2)/【(√6+√3)(√3+√2)】
=1/(√3+√2)+3/(√6+√3)
=(√3-√2)/【(√3+√2)(√3-√2)】+3(√6-√3)【(√6-√3)(√6+√3)】
=(√3-√2)+(√6-√3)
=√6-√2
[(n+1)√n--n*√(n+1)]/√(n+1)-√n
={√(n+1)√[(n+1)(n)]-√n√[(n+1)(n)]}/[√(n+1)-√n]
=√[(n+1)(n)]
[(n+1)√n--n*√(n+1)]/√(n+1)-√n
=[(n+1)√n-n√(n+1)]/[(√(n+1)-√n)]
=√[n*(n+1)] * [√(n+1)-√n]/[√(n+1)-√n]
=√[n*(n+1)]
一、(√6+4*√3+3*√2)/[(√6+√3)(√3+√2)]
=(√6+√3)+3(√3+√2)/[(√6+√3)(√3+√2)]
=1/(√3+√2)+3/(√6+√3)
=(√3-√2)/[(√3+√2)(√3-√2)]+3(√6-√3)/[(√6+√3)(√6-√3)]
=(√3-√2)+(√6-√3)
=√6-√2
二、[(...
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一、(√6+4*√3+3*√2)/[(√6+√3)(√3+√2)]
=(√6+√3)+3(√3+√2)/[(√6+√3)(√3+√2)]
=1/(√3+√2)+3/(√6+√3)
=(√3-√2)/[(√3+√2)(√3-√2)]+3(√6-√3)/[(√6+√3)(√6-√3)]
=(√3-√2)+(√6-√3)
=√6-√2
二、[(n+1)√n-n√(n+1)]/√(n+1)-√n
分子化简
(n+1)√n-n√(n+1)
提取公因式:√(n+1)*√n
=[√(n+1)*√n]*[√(n+1)-√n]
分子分母同时乘以[√(n+1)+√n]
分母=[√(n+1)-√n]*[√(n+1)+√n]
=(n+1)-n
=1
分子=[√(n+1)*√n]*[√(n+1)-√n]*[√(n+1)+√n]
=[√(n+1)*√n]*(n+1-n)
=√(n^2+n)
答案=√(n^2+n)
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