Question

решите уравнение(х-4)^2(х+10)=15(х-4)

Answer 1

$(x-4)^{2}(x+10)=15(x-4) (x-4)^{2}(x+10)-15(x-4)=0 (x-4)((x-4)(x+10)-15)=0 (x-4)( x^{2} +6x-40-15)=0 (x-4)( x^{2}+6x-55 )=0 x_{1}=4 sqrt{D} = sqrt{256}=16 x_{2}=5 x_{3}=-11$

Answer 2

$(x-4)^2(x+10)=15(x-4)\(x-4)^2(x+10)-15(x-4)=0\(x-4)((x-4)(x+10)-15)=0\(x-4)(x^2+6x-55)=0\x^2+6x-55=0 x-4=0\x_{1,2}=-3^+_-8 x_3=4\x_1=-11 x_2=5\OTBET:x=-11,x=4,x=5$