Știind că X supra 1 + 1 supra X = 4, calculati X² supra 1 + 1 supra X² și X³ supra 1 + 1 supra X³. Va rog sa ma ajutati!

[tex]\displaystyle\\x + \frac{1}{x} = 4 \ , \text{se inmulteste tot randul cu x} \\ \\x^2 + 1 = 4x \\ \\x^2 - 4x + 1 = 0 \\ \\\Delta = (-4)^2 - 4 * 1 = 16 - 4 = 12 \\ \\x_1 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \\ \\x_2 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{4-2\sqrt{3}}{2} = 2 - \sqrt{3} \\ \\[/tex]

[tex]\displaystyle\\a) \ \text{Daca} \ x = 2 + \sqrt{3} \ , \ x^2 = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4 \sqrt{3} \\ \\ analog \implies x^2 = 7 - 4\sqrt{3} \ daca \ x = 2 - \sqrt{3}[/tex]

[tex]\displaystyle\\Pentru \ x^2 = 7+4\sqrt{3} \ , \ x^2 + \frac{1}{x^2} = 7 + 4\sqrt{3} + \frac{1}{7 + 4\sqrt{3}} \\ \\= \frac{(7+4\sqrt{3})^2 + 1}{7+4\sqrt{3}} = \frac{49 + 56\sqrt{3} + 16 \times 3 + 1}{7+4\sqrt{3}} \\ \\= \frac{98 + 56\sqrt{3}}{7 + 4\sqrt{3}} = 14 \times \frac{7+4\sqrt{3}}{7+4\sqrt{3}} \\ \\\boxed{=14}[/tex]

[tex]\displaystyle\\Pentru \ x^2 = 7 - 4\sqrt{3} \ , \ x^2 + \frac{1}{x^2}=7-4\sqrt{3} + \frac{1}{7+4\sqrt{3}} \\ \\\text{Efectuezi calculele si vei obtine} \ x^2 + \frac{1}{x^2} = 14 - 8\sqrt{3}[/tex]

[tex]\displaystyle\\b) \ Pentru \ x=2+\sqrt{3} \ , \ x^3 = 26 + 15\sqrt{3} \implies x^3 + \frac{1}{x^3} = 52 \\ \\Pentru \ x = 2-\sqrt{3} \ , \ x^3 = 26 - 15\sqrt{3} \implies x^3 + \frac{1}{x^3} = 52 \\\\Deci \ x^3 + \frac{1}{x^3} \ \text{va avea aceeasi valoare pentru ambele radacini} \ x_1 \ si \ x_2[/tex]

Andyilye Andyilye · Answer 2 · 2022-09-09T23:33:37+03:00

Răspuns:

a) 14; b) 52

Explicație pas cu pas:

[tex]\frac{x}{1} + \frac{1}{x} = 4 \\ [/tex]

a)

[tex]{\Big(\frac{x}{1} + \frac{1}{x}\Big)}^{2} = \frac{ {x}^{2} }{1} + 2 \cdot \frac{x}{1} \cdot \frac{1}{x} + \frac{1}{ {x}^{2} } = \\ = \frac{ {x}^{2} }{1} + \frac{1}{ {x}^{2} } + 2[/tex]

[tex]{4}^{2} = \frac{ {x}^{2} }{1} + \frac{1}{ {x}^{2} } + 2 \\ \frac{ {x}^{2} }{1} + \frac{1}{ {x}^{2} } = 16 - 2 \\ \implies \frac{ {x}^{2} }{1} + \frac{1}{ {x}^{2} } = 14[/tex]

b)

[tex]{\Big(\frac{x}{1} + \frac{1}{x}\Big)}^{3} = \frac{ {x}^{3} }{1} + 3 \cdot \frac{ {x}^{2} }{1} \cdot \frac{1}{x} + 3 \cdot \frac{x}{1} \cdot \frac{1}{ {x}^{2} } + \frac{1}{ {x}^{3} } = \\ = \frac{ {x}^{3} }{1} + 3 \cdot \frac{x}{1} + 3 \cdot \frac{1}{x} + \frac{1}{ {x}^{3} } = \frac{ {x}^{3} }{1} + 3 \cdot\Big(\frac{x}{1} + \frac{1}{x}\Big) + \frac{1}{ {x}^{3} }[/tex]

[tex]{4}^{3} = \frac{ {x}^{3} }{1} + 3 \cdot 4 + \frac{1}{ {x}^{3} } \\ \frac{ {x}^{3} }{1} + \frac{1}{ {x}^{3} } = 64 - 12 \\ \implies \frac{ {x}^{3} }{1} + \frac{1}{ {x}^{3} } = 52[/tex]