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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: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
* Nếu x + y + z = 0
\(A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\dfrac{x+y}{x}\cdot\dfrac{y+z}{y}\cdot\dfrac{z+x}{z}=\dfrac{\left(-z\right)}{x}\cdot\dfrac{\left(-x\right)}{y}\cdot\dfrac{\left(-y\right)}{z}=\dfrac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-\dfrac{xyz}{xyz}=-1\)
* Nếu x + y + z khác 0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-y-z}{x}=\dfrac{y-x-z}{y}=\dfrac{-x-y+z}{z}=\dfrac{x-y-z+y-x-z-x-y+z}{x+y+z}=\dfrac{-x-y-z}{x+y+z}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x-y-z=-x\\y-x-z=-y\\-x-y+z=-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\Rightarrow x=y=z\)
\(\Rightarrow A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2\cdot2\cdot2=8\)
Ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}\)
\(\Leftrightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=-\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}.\left(-\dfrac{1}{z}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
\(\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3}{xyz}.xyz\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vậy...
Theo bài ra :
\(\dfrac{x^2-yz}{x\left(1-yz\right)}=\dfrac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\left(x^2-yz\right)\left(y-xyz\right)=\left(y^2-xz\right)\left(x-xyz\right)\)
\(\Leftrightarrow x^2y-x^3yz-y^2z+xx^2z^2=xy^2-xy^3z-x^2z+x^2yz^2=0\)
\(\Leftrightarrow xy\left(x-y\right)-xyz\left(x^2-y^2\right)+z\left(x^2-y^2\right)+xyz^2\left(x-x\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2\right)\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(xy+xz+yz-xyz\left(x+y+z\right)\right)=0\)
Mà theo đề bài :
\(x\ne y\Rightarrow xy+xz+yz-xyz\left(x+y+z\right)=0\)
\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)
\(\Leftrightarrow\dfrac{xy}{xyz}+\dfrac{xz}{xyz}+\dfrac{yz}{xyz}=\dfrac{xyz\left(z+y+x\right)}{xyz}\)
\(\Leftrightarrow\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}=x+y+z\left(đpcm\right)\)