Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 419

а) ($$\sqrt{x}$$ + 1)($$\sqrt{x}$$ − 1) = $$\sqrt{x}$$² − 1² = x − 1
б) ($$\sqrt{x}$$ − $$\sqrt{a}$$)($$\sqrt{x}$$ + $$\sqrt{a}$$) = $$\sqrt{x}$$² − $$\sqrt{a}$$² = x − a
в) ($$\sqrt{m}$$ + $$\sqrt{2}$$)² = $$\sqrt{m}$$² + 2$$\sqrt{m}$$ ⋅ $$\sqrt{2}$$ + $$\sqrt{2}$$² = m + 2$$\sqrt{m ⋅ 2}$$ + 2 = m + 2$$\sqrt{2m}$$ + 2
г) ($$\sqrt{3}$$ − $$\sqrt{x}$$)² = $$\sqrt{3}$$² − 2$$\sqrt{3}$$ ⋅ $$\sqrt{x}$$ + $$\sqrt{x}$$² = 3 − 2$$\sqrt{3x}$$ + x
д) (5$$\sqrt{7}$$ − 13)(5$$\sqrt{7}$$ + 13) = (5$$\sqrt{3}$$)² − 13² = 5² ⋅ $$\sqrt{7}$$² − 13² = 25 ⋅ 7 − 169 = 175 − 169 = 6
е) (2$$\sqrt{2}$$ + 3$$\sqrt{3}$$)(2$$\sqrt{2}$$ − 3$$\sqrt{3}$$) = (2$$\sqrt{2}$$)² − (3$$\sqrt{3}$$)² = 2² ⋅ $$\sqrt{2}$$² − 3² ⋅ $$\sqrt{3}$$² = 4 ⋅ 2 − 9 ⋅ 3 = 8 − 27 = −19
ж) (6 − $$\sqrt{2}$$)² + 3$$\sqrt{32}$$ = 6² − 2 ⋅ 6 ⋅ $$\sqrt{2}$$ + $$\sqrt{2}$$² + 3$$\sqrt{16 ⋅ 2}$$ = 36 − 12$$\sqrt{2}$$ + 2 + 3 ⋅ 4$$\sqrt{2}$$ = 38 − 12$$\sqrt{2}$$ + 12$$\sqrt{2}$$ = 38
з) ($$\sqrt{2}$$ + $$\sqrt{18}$$)² − 30 = $$\sqrt{2}$$² + 2$$\sqrt{2}$$ ⋅ 18 + $$\sqrt{18}$$² − 30 = 2 + 2$$\sqrt{2 ⋅ 18}$$ + 18 − 30 = −10 + 2$$\sqrt{36}$$ = −10 + 2 ⋅ 6 = −10 + 12 = 2