Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 154

а) a² − 25/a + 3 ⋅ 1/a² + 5a − a + 5/a² − 3a = (a − 5)(a + 5)/a + 3 ⋅ 1/a(a + 5) − a + 5/a² − 3a = (a − 5)(a + 5) ⋅ 1/(a + 3) ⋅ a(a + 5) − a + 5/a² − 3a = (a − 5)(a + 5) : (a + 5)/(a + 3) ⋅ a(a + 5) : (a + 5) − a + 5/a² − 3a = a − 5/a(a + 3) − a + 5/a(a − 3) = (a − 5) ⋅ (a − 3)/a(a + 3) ⋅ (a − 3) − (a + 5) ⋅ (a + 3)/a(a − 3) ⋅ (a + 3) = a² − 3a − 5a + 15 − a² − 3a − 5a − 15/a(a + 3)(a − 3) = −16a : a/a(a + 3)(a − 3) : a = −16/a² − 9 = 16/9 − a²
б) 1 − 2x/2x + 1 + x² + 3x/4x² − 1 : 3 + x/4x +2 = 1 − 2x/2x + 1 + x(x + 3)/(2x − 1)(2x + 1) : 3 + x/4x +2 = 1 − 2x/2x + 1 + x(x + 3)/(2x − 1)(2x + 1) ⋅ 2(2x + 1)/3 + x = 1 − 2x/2x + 1 + x(x + 3) ⋅ 2(2x + 1)/(2x − 1)(2x + 1) ⋅ (3 + x) = 1 − 2x/2x + 1 + x(x + 3) ⋅ 2(2x + 1) : (2x + 1) ⋅ (3 + x)/(2x − 1)(2x + 1) ⋅ (3 + x) : (2x + 1) ⋅ (3 + x) = 1 − 2x/2x + 1 + 2x/2x − 1 = (1 − 2x) ⋅ (2x − 1)/(2x + 1) ⋅ (2x − 1) + 2x ⋅ (2x + 1)/(2x − 1) ⋅ (2x + 1) = −(2x − 1) ⋅ (2x − 1) + 2x ⋅ (2x + 1)/(2x + 1)(2x − 1) = −(2x − 1)² + 2x² + 2x/(2x + 1)(2x − 1) = −(4x² − 4x + 1) + 4x² + 2x/(2x + 1)(2x − 1) = −4x² + 4x − 1 + 4x² + 2x/(2x)² − 1² = 6x − 1/4x² − 1
в) b − c/a + b − ab − b²/a² − ac ⋅ a² − c²/a² − b² = b − c/a + b − b(a − b)/a(a − c) ⋅ (a − c)(a + c)/(a − b)(a + b) = b − c/a + b − b(a − b) ⋅ (a − c)(a + c)/a(a − c) ⋅ (a − b)(a + b) = b − c/a + b − b(a − b) ⋅ (a − c)(a + c) : (a − b)(a − c)/a(a − c) ⋅ (a − b)(a + b) : (a − b)(a − c) = b − c/a + b − b(a + c)/a(a + b) = (b − c) ⋅ a/(a + b) ⋅ a − b(a + c)/a(a + b) = ab − ac − ab − bc/a(a + b) = −ac − bc/a(a + b) = −c(a + b)/a(a + b) = −c(a + b) : (a + b)/a(a + b) : (a + b) = −c/a = −c/a
г) a² − 4/x² − 9 : a² − 2a/xy + 3y + 2 − y/x − 3 = (a − 2)(a + 2)/(x − 3)(x + 3) : a(a − 2)/y(x + 3) + 2 − y/x − 3 = (a − 2)(a + 2)/(x − 3)(x + 3) ⋅ y(x + 3)/a(a − 2) + 2 − y/x − 3 = (a − 2)(a + 2) ⋅ y(x + 3)/(x − 3)(x + 3) ⋅ a(a − 2) + 2 − y/x − 3 = (a − 2)(a + 2) ⋅ y(x + 3) : (x + 3)(a − 2)/(x − 3)(x + 3) ⋅ a(a − 2) : (x + 3)(a − 2) + 2 − y/x − 3 = y(a + 2)/a(x − 3) + 2 − y/x − 3 = y(a + 2)/a(x − 3) + (2 − y) ⋅ a/(x − 3) ⋅ a = y(a + 2) + a(2 − y)/a(x − 3) = ay + 2y + 2a − ay/a(x − 3) = 2y + 2a/a(x − 3) = 2(y + a)/a(x − 3)