Toán Lớp 8: Bài 4: Tìm x a) $x^{2}$ +5x =0 b) 2x (x-3) +(x-3) =0 c) (x+2) (x-1) -x+1= 0 d) x+3 = $(x+3)^{2}$

Question

Toán Lớp 8:  Bài 4: Tìm x a)  $x^{2}$ +5x =0 b) 2x (x-3) +(x-3) =0 c) (x+2) (x-1) -x+1= 0 d) x+3 =  $(x+3)^{2}$

dantam45 · Answer

Lời giải : a)x^2+5x=0 <=>x(x+5)=0 <=> ( left[ begin{array}{l}x=0 x+5=0 end{array} right. ) <=> ( left[ begin{array}{l}x=0 x=-5 end{array} right. ) Vậy phương trình có tập nghiệm : S={0;-5} b)2x (x-3) +(x-3) =0 <=>(2x+1)(x-3)=0 <=> ( left[ begin{array}{l}2x+1=0 x-3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-1/2 x=3 end{array} right. ) Vậy phương trình có tập nghiệm : S={-1/2;3} c)(x+2) (x-1) -x+1= 0 <=>(x+2)(x-1)-(x-1)=0 <=>(x+2-1)(x-1)=0 <=>(x+1)(x-1)=0 <=> ( left[ begin{array}{l}x+1=0 x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-1 x=1 end{array} right. ) Vậy phương trình có tập nghiệm : S={+-1} d)x+3=(x+3)^2 <=>x+3=x^2+6x+9 <=>x^2+5x+6=0 <=>(x^2+3x)+(2x+6)=0 <=>x(x+3)+2(x+3)=0 <=>(x+2)(x+3)=0 <=> ( left[ begin{array}{l}x+2=0 x+3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-2 x=-3 end{array} right. ) Vậy phương trình có tập nghiệm : S={-2;-3}

thanhthuythu · Answer

a)   x^2 + 5x = 0   &rArr; x(x + 5) = 0   &rArr;  ( left[  begin{array}{l}x = 0  x = -5 end{array}  right. )    Vậy  S = {0;-5}   b)   2x(x - 3) + (x - 3) = 0   &rArr; (2x + 1)(x - 3) = 0   &rArr;  ( left[  begin{array}{l}2x + 1 = 0  x - 3 = 0 end{array}  right. )    &rArr;  ( left[  begin{array}{l}x = -1/2  x = 3 end{array}  right. )    Vậy  S = {-1/2 ; 3}   c)   (x + 2)(x - 1) - x + 1 = 0   &rArr; (x + 2)(x - 1) - (x - 1) = 0   &rArr; (x + 2 - 1)(x - 1) = 0   &rArr; (x + 1)(x - 1) = 0   &rArr;  ( left[  begin{array}{l}x = -1  x = 1 end{array}  right. )    Vậy  S = {-1 ; 1}   d)   x + 3 = (x + 3)^2   &rArr; (x + 3)^2 - (x + 3) = 0   &rArr; (x + 3)(x + 3 - 1) = 0   &rArr; (x + 3)(x + 2) = 0   &rArr;  ( left[  begin{array}{l}x = -3  x = -2 end{array}  right. )    Vậy  S = {-3 ; -2}