Toán Lớp 8: Tìm x X^3(x-3)-x+3=0 5x^3(x-3)-5x^2+15x=0 X^3(x-5)=x-5 4x^2(x-7)=-4x+28 5x(x-4)=-12+3x 5x^2-5x-2x^3+2x^4=0

Question

Toán Lớp 8:  Tìm x X^3(x-3)-x+3=0 5x^3(x-3)-5x^2+15x=0 X^3(x-5)=x-5 4x^2(x-7)=-4x+28 5x(x-4)=-12+3x 5x^2-5x-2x^3+2x^4=0

giahan44 · Answer

a)x^3(x-3)-x+3=0 <=>(x-3)(x^3-1)=0 <=> ( left[ begin{array}{l}x=3 x^3=1 end{array} right. ) <=> ( left[ begin{array}{l}x=3 x=1 end{array} right. ) b)5x^3(x-3)-5x^2+15x=0 <=>5x^3(x-3)-5x(x-3)=0 <=>(x-3)(5x^3-5x)=0 <=>5x(x-3)(x^2-1)=0 <=>5x(x-3)(x-1)(x+1)=0 <=>5x=0 hoặc x-3=0 hoặc x-1=0 hoặc x+1=0 <=>x=0 hoặc x=3 hoặc x=1 hoặc x=-1 c)x^3(x-5)=x-5 <=>x^3(x-5)-(x-5)=0 <=>(x-5)(x^3-1)=0 <=> ( left[ begin{array}{l}x=5 x^3=1 end{array} right. ) <=> ( left[ begin{array}{l}x=3 x=1 end{array} right. ) d)4x^2(x-7)=-4x+28 <=>4x^2(x-7)+4x-28=0 <=>4x^2(x-7)+4(x-7)=0 <=>4(x-7)(x^2+1)=0 <=>x-7=0(vì 4 và x^2+1>0) <=>x=7 e)5x(x-4)=-12+3x <=>5x(x-4)-3x+12=0 <=>5x(x-4)-3(x-4)=0 <=>(x-4)(5x-3)=0 <=> ( left[ begin{array}{l}x=4 5x=3 end{array} right. ) <=> ( left[ begin{array}{l}x=4 x= frac{3}{5} end{array} right. ) f)5x^2-5x-2x^3+2x^4=0 <=>5x(x-1)+2x^3(x-1)=0 <=>x(x-1)(5+2x^2)=0 <=> ( left[ begin{array}{l}x=0 x=1 end{array} right. ) (5+2x^2>0)

thanhbinh2k5 · Answer

a,  x&sup3;(x-3)-x+3=0   &hArr; x&sup3;(x-3)-(x-3)=0   &hArr; (x-3)(x-1)(x&sup2;+x+1)=0   &hArr; x=3 &or; x=1   b,  5x&sup3;(x-3)-5x&sup2;+15x=0   &hArr; 5x&sup3;(x-3)-5x(x-3)=0   &hArr; 5x(x-1)(x+1)(x-3)=0   &hArr; x=0 &or; x=1 &or; x=-1 &or; x=3   c,  x&sup3;(x-5)=x-5   &hArr; (x-5)(x-1)(x&sup2;+x+1)=0   &hArr; x=5 &or; x=1   d,  4x&sup2;(x-7)=-4x+28   &hArr; 4x&sup2;(x-7)+4(x-7)=0   &hArr; 4(x&sup2;+1)(x-7)=0   &hArr; x=7   e,  5x(x-4)=-12+3x   &hArr; 5x(x-4)-3(x-4)=0   &hArr; (x-4)(5x-3)=0   &hArr; x=4 &or; x=3/5   g, 5x&sup2;-5x-2x&sup3;+2x^4=0   &hArr; 5x(x-1)+2x&sup3;(x-1)=0   &hArr; x(2x&sup2;+5)(x-1)=0   &hArr; x=0 &or; x=1