Toán Lớp 8: Giải phương trình x(x+1)(x+2)(x+3) = 24.

Question

Toán Lớp 8:  Giải phương trình  x(x+1)(x+2)(x+3) = 24.

giahan44 · Answer

x(x+1)(x+2)(x+3)=24   =>x(x+3)(x+1)(x+2)=24   =>(x^2+3x)(x^2+2x+x+2)=24   =>(x^2+3x)(x^2+3x+2)=24   =>(x^2+3x+1-1)(x^2+3x+1+1)=24   Đặt x^2+3x+1=a   =>(a-1)(a+1)=24   =>a^2-1^2-24=0   =>a^2-25=0   =>a^2=25   => ( left[  begin{array}{l}a=5  a=-5 end{array}  right. )    Thay a=x^2+3x+1, ta c&oacute;:   => ( left[  begin{array}{l}x^2+3x+1=5  x^2+3x+1=-5 end{array}  right. )    => ( left[  begin{array}{l}x^2+3x-4=0  x^2+3x+6=0 end{array}  right. )    Ta c&oacute;:   x^2>=0&forall;x   =>x^2+3x+6=0 l&agrave; v&ocirc; l&yacute;   =>x^2+3x-4=0   =>x^2+4x-x-4=0   =>x(x+4)-(x+4)=0   =>(x+4)(x-1)=0   => ( left[  begin{array}{l}x+4=0  x-1=0 end{array}  right. )    => ( left[  begin{array}{l}x=-4  x=1 end{array}  right. )    Vậy S={-4;1}.