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救救我~ pls.....唔識做....(中二maths)

發問:

identity 1. ) x^2 - (x-1)^2 = 2x-1 2. ) (4x-1)/5 + (3x+2)/3 = (27x-7)/15 3.) [x(y-z)] / 2 + [ y(z-x)] /3 + [x(x-y)]/6 = (xy-2yz+yz)/ 6

最佳解答:

(1) L.H.S.=x^2-(x-1)^2 =x^2-(x^2-2x+1) =x^2-x^2+2x-1 =2x-1 R.H.S.=2x-1 becauseL.H.S.=R.H.S. ∴ x^2-(x-1)^2≡2x-1 (2) L.H.S.=(4x-1)/5+(3x+2)/3 =(12x-3)/15+(15x+10)/15 =(12x-3+15x+10)/15 =(27x+7)/15 because L.H.S.≠R.H.S. ∴ (4x-1)/5 + (3x+2)/3 = (27x-7)/15 is not an identity (3) L.H.S.=[x(y-z)]/2+[y(z-x)]/3+[x(x-y)]/6 =(xy-xz)/2+(yz-xy)/3+(x^2-xy)/6 =(3xy-3xz)/6+(2yz-2xy)/6+(x^2-xy)/6 =(3xy-3xz+2yz-2xy+x^2-xy)/6 =(-3xz+2yz+x^2)/6 R.H.S.=(xy-2yz+yz)/6 because L.H.S.≠R.H.S. ∴ [x(y-z)]/2+[y(z-x)]/3+[x(x-y)]/6=(xy-2yz+yz)/6 is not an identity.

其他解答:

You mean prove it? (1) L.H.S.=x^2-(x-1)^2 =x^2-(x^2-2x+1) =x^2-x^2+2x-1 =2x-1 R.H.S.=2x-1 ∴ L.H.S.=R.H.S. ∴ x^2-(x-1)^2≡2x-1 (2) L.H.S.=(4x-1)/5+(3x+2)/3 =3(4x-1)/15+5(3x+2)/15 =[3(4x-1)+5(3x+2)]/15 =(12x-3+15x+10)/15 =(12x+15x-3+10)/15 =(27x+7)/15 R.H.S.=(xy-2yz+yz)/6 ∴ L.H.S.≠R.H.S. ∴ (4x-1)/5 + (3x+2)/3 = (27x-7)/15 is not an identity, it is an equation. (3) L.H.S.=[x(y-z)]/2+[y(z-x)]/3+[x(x-y)]/6 =(xy-xz)/2+(yz-xy)/3+(x^2-xy)/6 =3(xy-xz)/6+2(yz-xy)/6+(x^2-xy)/6 =[3(xy-xz)+2(yz-xy)+(x^2-xy)]/6 =(3xy-3xz+2yz-2xy+x^2-xy)/6 =(3xy-2xy-xy-3xz+2yz+x^2)/6 =(-3xz+2yz+x^2) R.H.S.=(xy-2yz+yz)/6 ∴ L.H.S.≠R.H.S. ∴ [x(y-z)]/2+[y(z-x)]/3+[x(x-y)]/6=(xy-2yz+yz)/6 is not an identity. It is an equation.|||||(1) x2 - (x-1)2 = 2x-1 LHS =[x+(x-1)][x-(x-1)] LHS = (2x-1)(1) LHS= 2x-1 LHS = RHS (2) LHS = (4x-1)/5 + (3x+2)/3 LHS = [3(4x-1) + 5(3x+2)]/15 LHS = (12x-3+15x+10)/15 LHS = (27x+7)/15 RHS = (27x-7)/15 LHS ≠RHS (3) LHS = [x(y-z)] / 2 + [ y(z-x)] /3 + [x(x-y)]/6 LHS = [3x(y-z)] / 6 + [ 2y(z-x)] /6 + [x(x-y)]/6 LHS = (3xy-3xz)/6 + (2yz-2xy)/6 + (x2-xy)/6 LHS = (3xy-3xz+2yz-2xy+x2-xy)/6 LHS = (-3xz+2yz+x2)/6 RHS = (xy-2yz+yz)/ 6 LHS ≠RHS