Using properties of proportion solve:
k5+x5x5−k5=122121\dfrac{k^5 + x^5}{x^5 - k^5} = \dfrac{122}{121}x5−k5k5+x5=121122
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Given,
Applying componendo and dividendo we get,
⇒(k5+x5)+(x5−k5)(k5+x5)−(x5−k5)=122+121122−121⇒k5+x5+x5−k5k5+x5−x5+k5=2431⇒2x52k5=2431⇒(xk)5=(3)5⇒xk=3⇒x=3k.\Rightarrow \dfrac{(k^5 + x^5) + (x^5 - k^5)}{(k^5 + x^5) - (x^5 - k^5)} = \dfrac{122 + 121}{122 - 121} \\[1em] \Rightarrow \dfrac{k^5 + x^5 + x^5 - k^5}{k^5 + x^5 - x^5 + k^5} = \dfrac{243}{1} \\[1em] \Rightarrow \dfrac{2x^5}{2k^5} = \dfrac{243}{1} \\[1em] \Rightarrow \Big(\dfrac{x}{k}\Big)^5 = (3)^5 \\[1em] \Rightarrow \dfrac{x}{k} = 3 \\[1em] \Rightarrow x = 3k.⇒(k5+x5)−(x5−k5)(k5+x5)+(x5−k5)=122−121122+121⇒k5+x5−x5+k5k5+x5+x5−k5=1243⇒2k52x5=1243⇒(kx)5=(3)5⇒kx=3⇒x=3k.
Hence, x = 3k.
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Find x, if 16(a−xa+x)3=a+xa−x16\Big(\dfrac{a - x}{a + x}\Big)^3 = \dfrac{a + x}{a - x}16(a+xa−x)3=a−xa+x
If a : b :: c : d then prove that:
a : (a - b) :: c : (c - d).
x2−x+1x2+x+1=112(1−x)104(1+x)\dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{112(1 - x)}{104(1 + x)}x2+x+1x2−x+1=104(1+x)112(1−x)
If x ≠ y and x2−x+1y2−y+1=x2+x+1y2+y+1\dfrac{x^2 - x + 1}{y^2 - y + 1} = \dfrac{x^2 + x + 1}{y^2 + y + 1}y2−y+1x2−x+1=y2+y+1x2+x+1, prove that xy = 1. Use properties of proportion.
