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

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

\(\left\{{}\begin{matrix}\left(x^2-yz\right)\left(y+z\right)=x^2y+x^2z-y^2z-yz^2\\\left(y^2-xz\right)\left(x+z\right)=y^2x+y^2z-x^2z-xz^2\\\left(z^2-xy\right)\left(x+y\right)=z^2x+z^2y-x^2y-xy^2\end{matrix}\right.\)

Đa thức trên bằng 0

\(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)

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

\(=\dfrac{-x^2\left(y-z\right)-y^2\left(z-x\right)-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

Xét: \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

\(=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\)

\(\)\(=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)

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

\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)

\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)

Thêm dấu - đằng trc nữa suy ra bt có giá trị bằng 1 :P

a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)

b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

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

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

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

\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)

a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)

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

=0

c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

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

\(=\dfrac{1}{xyz}\)

a,

\(-\dfrac{x}{\left(x-y\right)\left(z-x\right)}-\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(z-x\right)\left(y-z\right)}\)

\(\dfrac{-x\left(y-z\right)-y\left(z-x\right)-z\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

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

= 0

Quy đồng tính bình thường.

\(A=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)\(=\dfrac{2x^2+2y^2+2z^2-2xy-2yz-2xz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\dfrac{2yz+2xz+2xy-2x^2-2y^2-2z^2}{ }\)

=0

\(A=\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+1+\dfrac{y^2-xz}{\left(y+z\right)\left(y+x\right)}+1+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}+1-3\)

Xét \(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+1=\dfrac{x^2-yz+x^2+xz+xy+yz}{\left(x+y\right)\left(x+z\right)}\)

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

Tương tự: \(\left\{{}\begin{matrix}\dfrac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+1=\dfrac{y}{y+z}+\dfrac{y}{y+x}\\\dfrac{z^2-xy}{\left(z+y\right)\left(z+x\right)}+1=\dfrac{z}{z+y}+\dfrac{z}{z+x}\end{matrix}\right.\)

Cộng vế với vế ta được:

\(A=\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{y+x}+\dfrac{y}{y+z}+\dfrac{z}{z+x}+\dfrac{z}{z+y}-3\)

\(A=\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}-3=1+1+1-3=0\)

\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\)

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

\(=\dfrac{1}{x-y}+\dfrac{1}{z-x}+\dfrac{1}{y-z}+\dfrac{1}{x-y}+\dfrac{1}{z-x}+\dfrac{1}{y-z}\)

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

Tham khảo:

Chứng minh \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)} \dfrac{z-x}{\left(y-z\right)\left(y-x\right)} \dfrac{... - Hoc24