a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{x-y+z}{2}\)

a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=a+b+c\)

Dat  (x-y)²+(y-z)²+(x-z)²=A

=(x+y)³+z³-3x²y-3xy²-3xyz / A

=(x+y+z).(x²+2xy+y²-xy-yz+z²)-3xy(x+y+z) / A

=(x+y+z).(x²+y²+z²-xy-yz-xz) /A

=2(x+y+z).(x²+y²+z²-xy-yz-xz) /2A 

=(x+y+z)[ (x²-2xy+y²)+(y²-2yz+z²)+(x²-2xz+z²) / 2A

=(x+y+z).[ (x-y}²+(y-z)²+(x-z)² ] /2A

=(x+y+z). A /2A

=x+y+z /2

kimh thế

nếu mẫu toàn cộng thì còn làm đc

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}=\frac{x+y+z}{2}\)

Biến đổi tử thức ta đc:

   x³ - y³ + z³ + 3xyz

= (x - y)³ + z³ + 3x²y - 3xy² + 3xyz

= (x - y + z) [ (x - y)² - (x - y)z + z² ]  +  3xy(x - y + z)

= (x - y + z)(x² - 2xy + y² - xz + yz + z² + 3xy)

= (x - y + z)(x² + y² + z² + xy + yz - xz)

Biến đổi mẫu thức ta đc:

     (x + y)² + (y + z)² + (z - x)²

= x² + 2xy + y² + y² + 2yz + z² + z² - 2xz + x²

= 2(x² + y² + z² + xy + yz - xz)

 Vậy  A = \(\frac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)\(=\frac{x-y+z}{2}\)