Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 876

а) ³$$\sqrt{5\sqrt{2} + 7}$$ – ³$$\sqrt{5\sqrt{2} – 7}$$ = ³$$\sqrt{2\sqrt{2} + 3\sqrt{2} + 6 + 1}$$ – ³$$\sqrt{2\sqrt{2} + 3\sqrt{2} - 6 - 1}$$ = ³$$\sqrt{(\sqrt{2})^2 + 3 · (\sqrt{2})^2 · 1 + 3 · \sqrt{2} · 1^2 + 1^3}$$ – ³$$\sqrt{(\sqrt{2})^2 - 3 · (\sqrt{2})^2 · 1 + 3 · \sqrt{2} · 1^2 - 1^3}$$ = ³$$\sqrt{(\sqrt{2} + 1)^3}$$ – ³$$\sqrt{(\sqrt{2} - 1)^3}$$ = $$\sqrt{2}$$ + 1 – ($$\sqrt{2}$$ – 1) = 2

б) ³$$\sqrt{2 + \sqrt{5}}$$ + ³$$\sqrt{2 – \sqrt{5}}$$

обозначим х = ³$$\sqrt{2 + \sqrt{5}}$$, у = ³$$\sqrt{2 – \sqrt{5}}$$, тогда

х³ + у³ = (³$$\sqrt{2 + \sqrt{5}}$$)³ + (³$$\sqrt{2 – \sqrt{5}}$$)³ = 2 + $$\sqrt{5}$$ + 2 - $$\sqrt{5}$$ = 4

x³ + y³ = (x + y)(x² – xy + y²)

ху = ³$$\sqrt{2 + \sqrt{5}}$$ · ³$$\sqrt{2 – \sqrt{5}}$$ = ³$$\sqrt{(2 + \sqrt{5})(2 - \sqrt{5})}$$ = ³$$\sqrt{2^2 - (\sqrt{5}^2)}$$ = ³$$\sqrt{4 - 5}$$ = ³$$\sqrt{-1}$$ = –1

(x + y)² = x² + 2xy + y²
x² + y² = (x + y)² – 2xy = (x + y)² – 2 · (–1) = (x + y)² + 2
4 = (x + y)(x² + y² – xy)
4 = (x + y)((x + y)² + 2 – (–1))
4 = (x + y)((x + y)² + 3)
пусть x + y = m
4 = m(m² + 3)
m³ + 3m – 4 = 0
m³ + 3m – 1 – 3 = 0
(m³ – 1) + (3m – 3) = 0
(m – 1)(m² + m + 1) + 3(m – 1) = 0
(m – 1)(m² + m + 1 + 3) = 0
(m – 1)(m² + m + 4) = 0
m – 1 = 0 или m² + m + 4 = 0
m = 1               D = 1² – 4 · 1 · 4 = –15
                         корней нет
x + y = 1
x + y = ³$$\sqrt{2 + \sqrt{5}}$$ + ³$$\sqrt{2 – \sqrt{5}}$$ = 1

Ответ: 1