1.решите уравнение: а) (x^2+3x)/(x-4)=x^2/(4-x) - вопрос №8783083123424533223 от nastyafominto 05.07.2022 16:52

Question

Алгебра: 1.решите уравнение: а) (x^2+3x)/(x-4)=x^2/(4-x) б) 5/(y+3)-3/y=(2-y)/(y^2+3y)

nastyafominto · Answer

А) x² + 3x / x - 4 = x² / 4 - x(x² + 3x) • (4 - x) = x² • (x - 4)(x² - x³ + 12x - 3x² - x³ + 4x² = 05x² - 2x³ + 12x = 0x • (5x - 2x² + 12) = 0x • (-2x² + 5x + 12) = 0x • (-2x² + 8x - 3x + 12) = 0x • (-2x • (x - 4) - 3 • (x - 4)) = 0x • (-2x - 3) • (x - 4) = 0x2 = 0x1 = -3/2b) 5/y+3 - 3/y = 2-y/y²+3 5/y+3 - 3/y - 2-y/y²+3y = 05/y+3 - 3/y - 2-y/y•(y+3) = 05y-3(y+3)-(2-y)/y•(y+3) = 05y-3y-9-2+y/y•(y+3) = 03y-11/y•(y+3) = 03y - 11 = 03y = 11y = 11/3y = 3 2/3...