Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 472

3у² = 12
у² = 4
у₁ = 2, у₂ = −2
х₁ = 2 · 2 = 4
х₂ = 2 · (−2) = −4

8у² = 12
у² = 12/8 = 3/2

у₁ = $$\sqrt{\frac{3}{2}}$$ = $$\frac{\sqrt{6}}{2}$$

у₂ = −$$\sqrt{\frac{3}{2}}$$ = −$$\frac{\sqrt{6}}{2}$$

x₁ = −3 · $$\sqrt{\frac{3}{2}}$$ = −$$\frac{3 \sqrt{3}}{\sqrt{2}}$$ = $$\frac{-3 \sqrt{6}}{2}$$

x₂ = −3 · (−$$\sqrt{\frac{3}{2}}$$) = $$\frac{3 \sqrt{3}}{\sqrt{2}}$$ = $$\frac{3 \sqrt{6}}{2}$$

Ответ: (4; 2), (−4; −2), ($$\frac{-3 \sqrt{6}}{2}$$; $$\frac{\sqrt{6}}{2}$$), ($$\frac{3 \sqrt{6}}{2}$$; −$$\frac{\sqrt{6}}{2}$$)

1) х – 3у = 0
х² – ху + у² = 7
х = 3у
(3у)² – 3у · у + у² = 7
9у² – 3у² + у² = 7
7у² = 7
у² = 1
у₁ = 1, у₂ = –1
х₁ = 3 · 1 = 3
х₂ = 3 · (–1) = –3

у² – 4у + 4 – у² + 2у + у² = 7
у² – 2у – 3 = 0

По теореме Виета
у₁ = 3, у₂ = –1
х₁ = 3 – 2 = 1
х₂ = –1 – 2 = –3

Ответ: (3; 1), (–3; –1), (1; 3)