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![](https://rs.olm.vn/images/avt/0.png?1311)
Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự thay vào mà quy đồng
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
\(x+2+\frac{3}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{x-2}+\frac{3}{x-2}\)
\(=\frac{x^2-4}{x-2}+\frac{3}{x-2}\)
\(=\frac{x^2-4+3}{x-2}\)
\(=\frac{x^2-1}{x-2}\)
2)
\(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2}{\left(x-y\right)\left(x-z\right)}-\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{y^2\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{\left(x^2-xy-xz+yz\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{x^2y-xy^2-xyz+y^2z-x^2z+xyz+xz^2-yz^2}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}\)
\(=1\)
S=1+4+7+..+n
Tổng S có số số hạng là \(\frac{\left(n-1\right)}{3}+1=\frac{n+2}{3}\)
Tổng S có giá trị là
\(S=\frac{\left(n+1\right)}{2}.\frac{n+2}{3}=\frac{\left(n+1\right)\left(n+2\right)}{6}\)