Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 705

а) $$\frac{21а^3 - 6а^2b}{12ab - 42a^2}$$ = $$\frac{3а^2(7a - 2b)}{6a(2b - 7a)}$$ = $$\frac{3а^2(7a - 2b)}{-6a(7a - 2b)}$$ = $$\frac{a}{-2}$$ = –$$\frac{a}{2}$$

б) $$\frac{6m^3 + 3mn^2}{2m^3n + mn^3}$$ = $$\frac{3m(2m^2 + n^2)}{mn(2m^2 + n^2)}$$ = $$\frac{3}{n}$$

в) $$\frac{x^2 - 2mx + 3x - 6m}{x^2 + 2mx + 3x + 6m}$$ = $$\frac{(x^2 - 2m) + (3x - 6m)}{(x^2 + 2mx) + (3x + 6m)}$$ = $$\frac{x(x - 2m) + 3(x - 2m)}{x(x + 2m) + 3(x + 2m)}$$ = $$\frac{x - 2m}{x + 2m}$$

г) $$\frac{8ab + 2a - 20b - 5}{4ab - 8b^2 + a - 2b}$$ = $$\frac{(8ab + 2a) - (20b + 5)}{(4ab - 8b^2) + (a - 2b)}$$ = $$\frac{2a(4b + 1) - 5(4b + 1)}{4ab(a - 2b) + (a - 2b)}$$ = $$\frac{(4b + 1)(2a - 5)}{(a - 2b)(4b + 1)}$$ = $$\frac{2a - 5}{a - 2b}$$

д) $$\frac{16a^2 - 8ab + b^2}{16a^2 - b^2}$$ = $$\frac{(4a - b)^2}{(4a - b)(4a + b)}$$ = $$\frac{4a - b}{4a + b}$$

е) $$\frac{9x^2 - 25y^2}{9x^2 + 30xy + 25y^2}$$ = $$\frac{(3x - 5y)(3x + 5y)}{(3x + 5y)^2}$$ = $$\frac{3x - 5y}{3x + 5y}$$

ж) $$\frac{a^2 - 3a}{a^2 + 3a - 18}$$

По теореме Виета
a₁ = 3, a₂ = –6
a² + 3a – 18 = (a – 3)(a + 6)

$$\frac{a^2 - 3a}{a^2 + 3a - 18}$$ = $$\frac{a(a - 3)}{(a - 3)(a + 6)}$$ = $$\frac{a}{a + 6}$$

з) $$\frac{4x^2 - 8x + 3}{4x^2 - 1}$$

4x² – 8x + 3 = 0
D = (–8)² – 4 · 4 · 3 = 64 – 48 = 16

x = $$\frac{8 ± \sqrt{16}}{2 \cdot 4}$$ = $$\frac{8 ± 4}{8}$$

x₁ = 0,5, x₂ = 1,5
4x² – 8x + 3 = 4(x – 0,5)(x – 1,5) = (2x – 1)(2x – 3)

$$\frac{4x^2 - 8x + 3}{4x^2 - 1}$$ = $$\frac{(2x - 1)(2x - 3)}{(2x - 1)(2x + 1)}$$ = $$\frac{2x - 3}{2x + 1}$$

и) $$\frac{m^2 + 4m - 5}{m^2 + 7m + 10}$$

m² + 4m – 5 = 0
По теореме Виета
m₁ = –5; m₂ = 1
m² + 4m – 5 = (m + 5)(m – 1)
m² + 7m + 10 = 0

По теореме Виета
m₁ = –2; m₂ = –5
m² + 7m + 10 = (m + 2)(m + 5)

$$\frac{m^2 + 4m - 5}{m^2 + 7m + 10}$$ = $$\frac{(m + 5)(m - 1)}{(m + 2)(m + 5)}$$ = $$\frac{m - 1}{m + 2}$$