救救我~ pls.....唔識做....(中二maths)

(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.

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