Evaluate: a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3\dfrac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times (a^2)^{2n+3}}an(4n−1)×(a2)2n+3a2n+1×a(2n+1)(2n−1)
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a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3=a2n+1×a4n2−1a4n2−n×a4n+6=a(2n+1)+(4n2−1)a(4n2−n)+(4n+6)=a2n+4n2a4n2+3n+6=a(2n+4n2)−(4n2+3n+6)=a2n+4n2−4n2−3n−6=a−n−6=1an+6\dfrac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times (a^2)^{2n+3}}\\[1em] = \dfrac{a^{2n+1}\times a^{4n^2-1}}{a^{4n^2-n}\times a^{4n+6}}\\[1em] = \dfrac{a^{(2n+1)+(4n^2-1)}}{a^{(4n^2-n)+(4n+6)}}\\[1em] = \dfrac{a^{2n+4n^2}}{a^{4n^2+3n+6}}\\[1em] = a^{(2n+4n^2)-(4n^2+3n+6)}\\[1em] = a^{2n+4n^2-4n^2-3n-6}\\[1em] = a^{-n-6}\\[1em] = \dfrac{1}{a^{n+6}}an(4n−1)×(a2)2n+3a2n+1×a(2n+1)(2n−1)=a4n2−n×a4n+6a2n+1×a4n2−1=a(4n2−n)+(4n+6)a(2n+1)+(4n2−1)=a4n2+3n+6a2n+4n2=a(2n+4n2)−(4n2+3n+6)=a2n+4n2−4n2−3n−6=a−n−6=an+61
a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3=1an+6\dfrac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times (a^2)^{2n+3}} = \dfrac{1}{a^{n+6}}an(4n−1)×(a2)2n+3a2n+1×a(2n+1)(2n−1)=an+61
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Show that: [xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=1\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a} = 1[x−bxa]a−b.[x−cxb]b−c.[x−axc]c−a=1
Evaluate: x5+n×(x2)3n+1x7n−2\dfrac{x^{5+n}\times(x^2)^{3n+1}}{x^{7n-2}}x7n−2x5+n×(x2)3n+1
Prove that: (m+n)−1(m−1+n−1)=(mn)−1(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}(m+n)−1(m−1+n−1)=(mn)−1
Prove that:
[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1[xbxa]ab1[xcxb]bc1[xaxc]ca1=1
