Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1282

$$ y=\sqrt{12+2 \sqrt{35+2 x-x^2}}-\sqrt{12-2 \sqrt{35+2 x-x^2}} $$
y² = $$ \left(\sqrt{12+2 \sqrt{35+2 x-x^2}}-\sqrt{12-2 \sqrt{35+2 x-x^2}}\right)^2 $$ = $$ =\left(\sqrt{12+2 \sqrt{35+2 x-x^2}}\right)^2- $$ $$ 2 \sqrt{12+2 \sqrt{35+2 x-x^2}} \sqrt{12-2 \sqrt{35+2 x-x^2}} $$ $$ =+\left(\sqrt{12-2 \sqrt{35+2 x-x^2}}\right)^2= $$ = 12 + 2$$\sqrt{35 + 2x − x²}$$ − $$ 2 \sqrt{\left(12+2 \sqrt{35+2 x-x^2}\right)\left(12-2 \sqrt{35+2 x-x^2}\right)} $$ + 12 − 2$$\sqrt{35 + 2x − x²}$$ = 24 − 2$$\sqrt{144 − 4(35 + 2x − x²)}$$ = 24 − 2$$\sqrt{144 − 140 − 8x + 4x²)}$$ = 24 − 2$$\sqrt{4 − 8x + 4x²)}$$ = 24 − 2$$\sqrt{4(x² − 2x + 1)}$$ = 24 − 2 · 2$$\sqrt{(x − 1)²}$$ = 24 − 4|x − 1|
y = $$\sqrt{24 − 4|x − 1|}$$
Если х − 1 ≥ 0, х ≥ 1, то y = $$\sqrt{24 − 4x + 4}$$ = $$\sqrt{28 − 4x}$$ x = 6, y = $$\sqrt{28 − 4 · 6}$$ = $$\sqrt{4}$$ = 2 x = 7, y = $$\sqrt{28 − 4 · 7}$$ = 0
Если х − 1 < 0, х < 1, то y = $$\sqrt{24 + 4x − 4}$$ = $$\sqrt{20 + 4x}$$ x = −1, y = $$\sqrt{20 + 4 · (−1)}$$ = $$\sqrt{16}$$ = 4 x = −5, y = $$\sqrt{20 + 4 · (−5)}$$ = 0
y = 0; 2; 4