Toán Lớp 8: tìm x x(x-5)-x+5=0 x^2-9=-2x(x-3)

Question

Toán Lớp 8:  tìm x x(x-5)-x+5=0 x^2-9=-2x(x-3)

kymmailien · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    x(x-5)-x+5 =0    &hArr; x(X-5) -1.(x-5) = 0   &hArr;(x-5)(x-1)=0   &hArr; ( left[  begin{array}{l}x-5=0  x-1=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=5  x=1 end{array}  right. )    vậy x= 1 hoặc x= 5   x^2 -9 =  -2x(x-3 )   &hArr; x^2 -3^2 +2x(x-3) =0    &hArr; (x-3)(x+3) +2x(x-3) = 0   &hArr; (x-3)(x+3+2x) = 0   &hArr; (x-3)(3x+3) = 0   &hArr; ( left[  begin{array}{l}x-3=0  3x+3=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=3  3x=-3 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=3  x=-1 end{array}  right. )    vậy x= -1 hoặc x= 3

hienthucthu97 · Answer

a) x(x-5)-x+5=0   &hArr;x(x-5)-(x-5)=0   &hArr;(x-5)(x-1)=0   &hArr; ( left[  begin{array}{l}x-5=0  x-1=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=5  x=1 end{array}  right. )    Vậy S={5,1}   b) x^2-9=-2x(x-3)   &hArr;x^2-9+2x(x-3)=0   &hArr;(x-3)(x+3)+2x(x-3)=0   &hArr;(x-3)(x+3+2x)=0   &hArr;(x-3)(3x+3)=0   &hArr;3(x-3)(x+1)=0   &hArr; ( left[  begin{array}{l}x-3=0  x+1=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=3  x=-1 end{array}  right. )    Vậy S={3,-1}