a) (x + 5).(x-4)= 0;
=> x+5 =0 => x= -5
    X-4=0   => x=4   
c) 25-(3+x)² = 3.(-2)³;
25-(3+x)²+24=0
49-9-6x-x^2=0
40-6x-x^2=0
40-10x+4x-x^2=0
(10+x)(4-x)=0
=>(10+x)=0 =>x=-10
=>(4-x)=0 =>x=4

b) (x² + 1).(x - 1) = 0;
=>(x² + 1)=0 => (x² =- 1( loại)
=>(x - 1)=0  => x=1
 

a) (x + 5).(x-4)= 0
TH1 : x+5 = 0
          x =-5
TH2: x-4=0
           x =4
c) 25-(3+x)² = 3.(-2)³
    25-(3+x)^2 = 3.(-8)
         (3+x)^2= 25+24
          (3+x)^2 = 49
TH1 : 3+x=7
              x =4
Th2 : 3+x=-7
            x=-10
b) (x² + 1).(x - 1) = 0;
        x-1=0 ( vì x^2+1 luôn lớn hơn 0 )
        x =1