đĨ² + šâđĨ² = ā§Š āĻšāĻ˛ā§,(āĻ) (đĨ + šâđĨ)² āĻāĻ° āĻŽāĻžāĻ¨ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĨ¤(āĻ) (đĨâļ + đŖ)âđĨÂŗ āĻāĻ° āĻŽāĻžāĻ¨ āĻāĻ¤?(āĻ) (đĨ² â šâđĨ²)Âŗ à đ āĻāĻ° āĻŽāĻžāĻ¨ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĨ¤
Created: 9 months ago
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Updated: 7 months ago
Asked by
XÃŖdiza
XÃŖdiza
(āĻ)
\( x + \frac{1}{x} = \sqrt{x² + 2 + \frac{1}{x²}} \)
\( x² + \frac{1}{x²} = ā§Š \)
āĻ¤āĻžāĻšāĻ˛ā§,
\( (x + \frac{1}{x})² = ā§Š + ā§¨ = ā§Ģ \)
āĻŽāĻžāĻ¨: ā§Ģ
(āĻ)
\( \frac{x^6 + 1}{x^3} = x^3 + \frac{1}{x^3} \)
\( (x + \frac{1}{x})² = ā§Ģ \) āĻĨā§āĻā§,
\( x + \frac{1}{x} = \sqrt{5} \)
āĻ¤āĻžāĻšāĻ˛ā§,
\( x^3 + \frac{1}{x^3} = (x + \frac{1}{x})(x² - 1 + \frac{1}{x²}) \)
āĻāĻāĻŋ āĻ§āĻ°ā§āĻ¨, \( x² + \frac{1}{x²} = ā§Š \)
āĻ¤āĻžāĻšāĻ˛ā§,
\( x^3 + \frac{1}{x^3} = (\sqrt{5})(ā§¨) = ā§¨\sqrt{5} \)
āĻŽāĻžāĻ¨: \( ā§¨\sqrt{5} \)
(āĻ)
\( (x² - \frac{1}{x²})Âŗ = ((x + \frac{1}{x})² - ā§¨)Âŗ = (ā§Ģ - ā§¨)Âŗ = ÛŗÂŗ = ā§¨ā§ \)
āĻāĻŦāĻ,
āĻāĻāĻŋ \( c \)-āĻāĻ° āĻ¸āĻžāĻĨā§ āĻā§āĻŖāĻŋāĻ¤ āĻšāĻ˛ā§,
āĻŽāĻžāĻ¨: \( ā§¨ā§c \)
AI-āĻāĻ° āĻāĻ¤ā§āĻ¤āĻ°ā§ āĻā§āĻ˛ āĻĨāĻžāĻāĻ¤ā§ āĻĒāĻžāĻ°ā§āĨ¤
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