Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 338

а) х³ + $$\frac{1}{x^3}$$ = 22(х + $$\frac{1}{x}$$)

х ≠ 0

t = x + 1/x

t³ = (x + 1/x)³ = x³ + 3x² · 1/x + 3x · (1/x)² + (1/x)³

t³ = x³ + 3x + 3 · 1/x + $$\frac{1}{x^3}$$

t³ = x³ + 3(x + 1/x) + $$\frac{1}{x^3}$$

t³ = x³ + 3 · t + $$\frac{1}{x^3}$$

x³ + $$\frac{1}{x^3}$$ = t³ − 3t

t³ – 3t = 22t
t³ – 3t – 22t = 0
t³ – 25t = 0
t(t² – 25) = 0
t(t – 5)(t + 5) = 0

t = 0
x + 1/x = 0
x² + 1 = 0
x² = –1
нет корней

или
t – 5 = 0
t = 5
x + 1/x = 5
x² + 1 = 5x
x² – 5x + 1 = 0
D = (–5)² – 4 = 21

x = $$\frac{5 ± \sqrt{21}}{2}$$

x₁ = $$\frac{5 + \sqrt{21}}{2}$$

x₂ = $$\frac{5 - \sqrt{21}}{2}$$

или
t + 5 = 0
t = –5
x + 1/x = 5
x² + 1 = –5x
x² + 5x + 1 = 0
D = 5² – 4 = 21

x = $$\frac{-5 ± \sqrt{21}}{2}$$

x₃ = $$\frac{-5 + \sqrt{21}}{2}$$

x₄ = $$\frac{-5 - \sqrt{21}}{2}$$

Ответ: $$\frac{5 + \sqrt{21}}{2}$$; $$\frac{5 - \sqrt{21}}{2}$$; $$\frac{-5 + \sqrt{21}}{2}$$; $$\frac{-5 - \sqrt{21}}{2}$$

б) х³ – $$\frac{1}{x^3}$$ = 19(х – 1/x)

х ≠ 0

t = x –1/x

t³ = (x – 1/x)³ = х³ – 3x² · 1/x + 3x · (1/x)² – (1/x)³

t³ = x³ – 3x + 3 · 1/x – $$\frac{1}{x^3}$$

t³ = x³ – 3(x – 1/x) – $$\frac{1}{x^3}$$

t³ = x³ – 3 · t – $$\frac{1}{x^3}$$

x³ – $$\frac{1}{x^3}$$ = t³ + 3t

t³ + 3t = 19t
t³ + 3t – 19t = 0
t³ – 16t = 0
t(t² – 16) = 0
t(t – 4)(t + 4) = 0

t = 0
x – 1/x = 0
x² – 1 = 0
x² = 1

x₁ = 1

x₂ = –1

или
t – 5 = 0
t = 5
x – 1/x = 4
x² – 1 = 4x
x² – 4x – 1 = 0
D = 4 + 1 = 5

x = $$\frac{2 ± \sqrt{5}}{1}$$ = 2 ± $$\sqrt{5}$$

x₃ = 2 + $$\sqrt{5}$$

x₄ = 2 – $$\sqrt{5}$$

или
t + 5 = 0
t = –5
x + 1/x = 5
x² + 1 = –5x
x² + 5x + 1 = 0
D = 5² – 4 = 21

x = $$\frac{-2 ± \sqrt{5}}{1}$$ = –2 ± $$\sqrt{5}$$

x₅ = –2 ± $$\sqrt{5}$$

x₆ = –2 – $$\sqrt{5}$$

Ответ: 1; –1; 2 + $$\sqrt{5}$$; 2 – $$\sqrt{5}$$; –2 ± $$\sqrt{5}$$; –2 – $$\sqrt{5}$$