Toán Lớp 8: CMR : `[2/(3x)-2/(x+1) .((x+1)/(3x)-x-1)]:(x-1)/x=(-2x)/(1-x)`

Question

Toán Lớp 8:  CMR :  [2/(3x)-2/(x+1) .((x+1)/(3x)-x-1)]:(x-1)/x=(-2x)/(1-x)

truonganhthi · Answer

Giải đáp + Lời giải và giải thích chi tiết:   C&oacute;:    VT= [ 2/(3x) - 2/(x+1) . ((x+1)/(3x) - x- 1 ) ] : (x-1)/x   VT= [ 2/(3x) - 2/(x+1) . ((x+1)/(3x) - (3x&sup2;)/(3x) - (3x)/(3x))]:(x-1)/x   VT= [ 2/(3x) - 2/(x+1). ((x+1-3x&sup2;-3x)/(3x) ) ] : (x-1)/x   VT= [ 2/(3x) - 2/(x+1) . (-3x&sup2;-2x+1)/(3x) ] : (x-1)/x   VT= [ 2/(3x) - 2/(x+1) . (-3x&sup2;-3x+x+1)/(3x)] : (x-1)/x   VT= [ 2/(3x) - 2/(x+1) . (-3x(x+1)+(x+1))/(3x)] : (x-1)/x   VT= [2/(3x) - (2(x+1)(1-3x))/((x+1).3x) ] : (x-1)/x   VT= [ 2/(3x) - (-6x&sup2;-4x+2)/(3x(x+1)] : (x-1)/x   VT= [(2(x+1))/(3x(x+1)) - (-6x&sup2;-4x+2)/(3x(x+1))] . x/(x-1)   VT= [(2x+2+6x&sup2;+4x-2)/(3x(x+1))] . x/(x-1)   VT= (6x&sup2;+6x)/(3x(x+1)) . x/(x-1)   VT= (2x(x+1))/((x+1)(x-1))   VT= (2x)/(x-1)   VT= (-2x)/(1-x)  (đpcm)

tonhu12 · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    [2/[3x] -2/[x+1] .([x+1]/[3x] -x-1)]:[x-1]/x   =[2/[3x]-2/[x+1]. [x+1-3x(x+1)]/[3x]]:[x-1]/x   =[2/[3x]-2/[x+1].[-3x^2 -2x+1]/[3x]]. x/[x-1]   =[ 2/[3x] -2/[x+1].[(x+1)(1-3x)]/[3x]]. x/[x-1]   =[2/[3x]-[2(1-3x)]/[3x] ]. x/[x-1]   =[2-2+6x]/[3x] . x/[x-1]   =[6x]/[3x] . x/[x-1]   = 2. x/[x-1]   =[2x]/[x-1]   =[-2x]/[1-x]   =>VT=VP(đpcm)   (B&agrave;i tham khảo)