Toán Lớp 8: Giúp mình với Câu 1 a) 6x(5x-2)-5x+2=0 b) 5(x+3)-2x(3+x)=0 c) 4x(x-2004)-x+2004=0 d) (x+1)²=x+1 e) x+3x-18=0 Mình cảm ơn ạ

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Toán Lớp 8:  Giúp mình với Câu 1 a) 6x(5x-2)-5x+2=0 b) 5(x+3)-2x(3+x)=0 c) 4x(x-2004)-x+2004=0 d) (x+1)²=x+1 e) x+3x-18=0 Mình cảm ơn ạ

cobebongdem · Answer

Hướng dẫn trả lời:       C&acirc;u 1:     a) 6x*(5x - 2) - 5x + 2 = 0   &harr; 6x*(5x - 2) - (5x - 2) = 0   &harr; (6x - 1)*(5x - 2) = 0   &harr; [(6x - 1 = 0), (5x - 2 = 0):}   &harr; [(6x = 1), (5x = 2):}   &harr; [(x = 1/6), (x = 2/5):}   Vậy x = 1/6 hoặc x = 2/5   b) 5*(x + 3) - 2x*(3 + x) = 0   &harr; 5*(x + 3) - 2x*(x + 3) = 0   &harr; (5 - 2x)*(x + 3) = 0   &harr; [(5 - 2x = 0), (x + 3 = 0):}   &harr; [(2x = 5), (x = - 3):}   &harr; [(x = 5/2), (x = - 3):}   Vậy x = 5/2 hoặc x = -3   c) 4x*(x - 2004) - x + 2004 = 0   &harr; 4x*(x - 2004) - (x - 2004) = 0   &harr; (4x - 1)*(x - 2004) = 0   &harr; [(4x - 1 = 0), (x - 2004 = 0):}   &harr; [(4x = 1), (x = 2004):}   &harr; [(x = 1/4), (x = 2004):}   Vậy x = 1/4 hoặc x = 2004   d) (x + 1)^2 = x + 1   &harr; (x + 1)^2 - (x + 1) = 0   &harr; (x + 1)*[(x + 1) - 1] = 0   &harr; (x + 1)*(x + 1 - 1) = 0   &harr; x*(x + 1) = 0   &harr; [(x = 0), (x + 1 = 0):}   &harr; [(x = 0), (x = - 1):}   Vậy x = 0 hoặc x = -1   e) Đề sai n&ecirc;n m&igrave;nh sửa nh&eacute;!   - Đề sửa: x^2 + 3x - 18 = 0   &harr; x^2 - 3x + 6x - 18 = 0   &harr; (x^2 - 3x) + (6x - 18) = 0   &harr; x*(x - 3) + 6*(x - 3) = 0   &harr; (x + 6)*(x - 3) = 0   &harr; [(x + 6 = 0), (x - 3 = 0):}   &harr; [(x = - 6), (x = 3):}   Vậy x = -6 hoặc x = 3

truongthanhhung · Answer

Giải đáp: a) S={2/5;1/6} b) S={-3;5/2} c) S={2004;1/4} d) S={-1;0} e) S={3;-6} Lời giải và giải thích chi tiết: a) 6x(5x-2)-5x+2=0 <=>6x(5x-2)-(5x-2)=0 <=>(5x-2)(6x-1)=0 <=> ( left[ begin{array}{l}5x-2=0 6x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}5x=2 6x=1 end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{2}{5} x= dfrac{1}{6} end{array} right. ) Vậy: S={2/5;1/6} b) 5(x+3)-2x(3+x)=0 <=>(x+3)(5-2x)=0 <=> ( left[ begin{array}{l}x+3=0 5-2x=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 -2x=-5 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 x= dfrac{5}{2} end{array} right. ) Vậy: S={-3;5/2} c) 4x(x-2004)-x+2004=0 <=>4x(x-2004)-(x-2004)=0 <=>(x-2004)(4x-1)=0 <=> ( left[ begin{array}{l}x-2004=0 4x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=2004 4x=1 end{array} right. ) <=> ( left[ begin{array}{l}x=2004 x= dfrac{1}{4} end{array} right. ) Vậy: S={2004;1/4} d) (x+1)^2=x+1 <=>(x+1)^2-(x+1)=0 <=>(x+1)(x+1-1)=0 <=>(x+1)x=0 <=> ( left[ begin{array}{l}x+1=0 x=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-1 x=0 end{array} right. ) Vậy: S={-1;0} e) Sửa đề: x^2+3x-18=0 <=>x^2-3x+6x-18=0 <=>x(x-3)+6(x-3)=0 <=>(x-3)(x+6)=0 <=> ( left[ begin{array}{l}x-3=0 x+6=0 end{array} right. ) <=> ( left[ begin{array}{l}x=3 x=-6 end{array} right. ) Vậy: S={3;-6}