X by x-y x-z + y by y-z y-x + x by z-x z-y
Answers
Answer:
(x−y)(x−z)
x
+
(y−z)(y−x)
y
+
(z−x)(z−y)
z
\mathsf{=\dfrac{x}{-(x-y)(z-x)}+\dfrac{y}{-(y-z)(x-y)}+\dfrac{z}{-(z-x)(y-z)}}=
−(x−y)(z−x)
x
+
−(y−z)(x−y)
y
+
−(z−x)(y−z)
z
\mathsf{=\dfrac{-x}{(x-y)(z-x)}-\dfrac{y}{(y-z)(x-y)}-\dfrac{z}{(z-x)(y-z)}}=
(x−y)(z−x)
−x
−
(y−z)(x−y)
y
−
(z−x)(y−z)
z
\textsf{Making the denominators equal}Making the denominators equal
\mathsf{=\dfrac{-x(y-z)-y(z-x)-z(x-y)}{(x-y)(y-z)(z-x)}}=
(x−y)(y−z)(z−x)
−x(y−z)−y(z−x)−z(x−y)
\mathsf{=\dfrac{-xy+xz-yz+xy-zx+yz}{(x-y)(y-z)(z-x)}}=
(x−y)(y−z)(z−x)
−xy+xz−yz+xy−zx+yz
\mathsf{=\dfrac{0}{(x-y)(y-z)(z-x)}}=
(x−y)(y−z)(z−x)
0
\mathsf{=0}=0
\implies\boxed{\mathsf{\dfrac{x}{(x-y)(x-z)}+\dfrac{y}{(y-z)(y-x)}+\dfrac{z}{(z-x)(z-y)}=0}}⟹
(x−y)(x−z)
x
+
(y−z)(y−x)
y
+
(z−x)(z−y)
z
=0