(5x + 2)(2x - 7) = 0; y ^ 2 - 5y + 2(y - 5) = 0; 5 - Dicamon

Question

(5x + 2)(2x - 7) = 0; y ^ 2 - 5y + 2(y - 5) = 0; 5/(x + 2) + 3/(x - 1) = (3x + 4)/((x + 2)(x - 1)); (x - 1)/(x + 1) - (x + 1)/(x - 1) = 8/(x ^ 2 - 1)

Answer

s(5x + 2)(2x - 7) = 0; G Gamma^ 5 (x+2=0; x = 2/5; 8x - 7 = 0; x = 9/2; y ^ 2 - 5y + 2(y - 5) = 0; Theta * y ^ 2 - 5y + 2y - 10 = 0; - y ^ 2 - 3y - 10 ^ u = 0; - (x + 2) * (x - 5)

Answer

(5x + 2)(2x - 7) = 0 0 - 2x - 7 = c = 7/2 5 0 y ^ 2 - 5y + 2y - 10 = c 3y 10 0 4 ^ 2 - 54; (=)+ 2 5 ^ 2 y=5; u = - 2 5 3x + 5 + 9 ( 2 ) 1 ) 3 2 ( + ) - ) ( 5(1 - 1) + 3(x + 2) - 3x

Giải chi tiết cho mình câu này với ạ.