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Problem: solve
${{6x}^{2}}$ – 25x + 12 + ${\frac{25}{x}}$ + ${\frac{6}{x^2}}$ = 0.
Solution: ${{6x}^{2}}$ – 25x + 12 + ${\frac{25}{x}}$ + ${\frac{6}{x^2}}$ = 0
= 6( ${x^2}$ + ${\frac{1}{x^2}}$) – 25(x – ${\frac{1}{x}}$) +12 = 0
= 6 (x – ${\frac{1}{x}})^2$ – 25 (x – ${\frac{1}{x}}$) + 24 = 0
= (x – ${\frac{1}{x}}$) = ${\frac{25\pm {\sqrt49}}{12}}$ = ${\frac{3}{2}}$ or ${\frac{8}{3}}$
If (x – ${\frac{1}{x}}$) = ${\frac{3}{2}}$
Or ${{2x}^2}$ – 3x – 2 = 0
Or x = 2 , – ${\frac{1}{2}}$
If (x – ${\frac{1}{x}}$) = ${\frac{8}{3}}$
Or ${x^2}$ – 1 = ${\frac{8}{3x}}$
Or ${3x^2}$ – 8x – 3 = 0
Or x = 3, -1/3
X = 2,- ${\frac{1}{2}}$ ,3,- ${\frac{1}{3}}$