Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 336

а) (х² + $$\frac{1}{x^2}$$ − (х + $$\frac{1}{x}$$) = 2

х ≠ 0

Пусть х + 1/х = t, тогда

t² = (x + $$\frac{1}{x}$$)² = x² + 2 · x · $$\frac{1}{x}$$ + $$\frac{1}{x^2}$$ = x² + 2 + $$\frac{1}{x^2}$$

x² + $$\frac{1}{x^2}$$ = t² − 2

2(t² – 2) – t = 2
2t² – 4 – t – 2 = 0
2t² – t – 6 = 0
D = (–1)² – 4 · 2 · (–6) = 49

t = $$\frac{1 ± \sqrt{49}}{2 \cdot 2}$$ = 1 ± 7/4

t₁ = 2, t₂ = –1,5

x + 1/x = 2     или     x + 1/x = –1,5
х² + 1 = 2х                 х² + 1 = –1,5х
х² – 2х + 1 = 0           х² + 1,5х + 1 = 0
(х – 1)² = 0                 D = 1,5² – 4 · 1 · 1 = –1,75

х = 1                            корней нет

Ответ: 1

б) 9х² – 18х + $$\frac{9}{x^2}$$ – $$\frac{18}{x}$$ = 22

9х² + $$\frac{9}{x^2}$$ – 18х – $$\frac{18}{x}$$ = 22

9(х² + $$\frac{1}{x^2}$$) – 18(х + $$\frac{1}{x}$$) = 22

х ≠ 0
Пусть х + = t, тогда

t² = (x + $$\frac{1}{x}$$)² = x² + 2 · x · $$\frac{1}{x}$$ + $$\frac{1}{x^2}$$ = x² + 2 + $$\frac{1}{x^2}$$

x² + $$\frac{1}{x^2}$$ = t² – 2

9(t² – 2) – 18t = 22
9t² – 18 – 18t – 22 = 0
9t² – 18t – 40 = 0
D = (–9)² – 9 · (–40) = 81 + 360 = 441

t = $$\frac{9 ± \sqrt{441}}{9}$$ = 9 ± 21/9

t₁ = 30/9 = 10/3 = 31/3

t₂ = –12/9 = –4/3 = –11/3

x + 1/x = 10/3     или      х + 1/х = –4/3
3х² + 3 = 10х                    3х² + 1 = –1,5х
3х² – 10х + 3 = 0              3х² + 4х + 3 = 0
D₁ = (–5)² – 3 · 3 = 16       D₁ = 2² – 3 · 3 = –5

х = $$\frac{5 ± \sqrt{16}}{3}$$ = 5 ± 4/3       нет корней
х₁ = 3, х₂ = 1/3

Ответ: 3 и 1/3