Simplify the following:
11+am−n+11+an−m\dfrac{1}{1 + a^{m - n}} + \dfrac{1}{1 + a^{n - m}}1+am−n1+1+an−m1
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Given,
⇒11+am−n+11+an−m⇒11+am.a−n+11+an.a−m⇒11+aman+11+anam⇒1an+aman+1am+anam⇒anan+am+amam+an⇒an+aman+am⇒1.\Rightarrow \dfrac{1}{1 + a^{m - n}} + \dfrac{1}{1 + a^{n - m}} \\[1em] \Rightarrow \dfrac{1}{1 + a^m.a^{-n}} + \dfrac{1}{1 + a^n.a^{-m}} \\[1em] \Rightarrow \dfrac{1}{1 + \dfrac{a^m}{a^n}} + \dfrac{1}{1 + \dfrac{a^n}{a^m}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{a^n + a^m}{a^n}} + \dfrac{1}{\dfrac{a^m + a^n}{a^m}} \\[1em] \Rightarrow \dfrac{a^n}{a^n + a^m} + \dfrac{a^m}{a^m + a^n} \\[1em] \Rightarrow \dfrac{a^n + a^m}{a^n + a^m} \\[1em] \Rightarrow 1.⇒1+am−n1+1+an−m1⇒1+am.a−n1+1+an.a−m1⇒1+anam1+1+aman1⇒anan+am1+amam+an1⇒an+aman+am+anam⇒an+aman+am⇒1.
Hence, 11+am−n+11+an−m\dfrac{1}{1 + a^{m - n}} + \dfrac{1}{1 + a^{n - m}}1+am−n1+1+an−m1 = 1.
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(xax−b)a2−ab+b2.(xbx−c)b2−bc+c2.(xcx−a)c2−ca+a2\Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2}.\Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2}.\Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2}(x−bxa)a2−ab+b2.(x−cxb)b2−bc+c2.(x−axc)c2−ca+a2
(a-1 + b-1) ÷ (a-2 - b-2)
Prove the following:
(a + b)-1(a-1 + b-1) = 1ab\dfrac{1}{ab}ab1
x+y+zx−1y−1+y−1z−1+z−1x−1=xyz\dfrac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} = xyzx−1y−1+y−1z−1+z−1x−1x+y+z=xyz
