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\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
1/
\(\dfrac{\left(x-y\right)^3-3xy\left(x+y\right)+y^3}{x-6y}\)
\(=\dfrac{x^3-3x^2y+3xy^2-y^3-3x^2y-3xy^2+y^3}{x-6y}\)
\(=\dfrac{x^3-6x^2y}{x-6y}\)
\(=\dfrac{x^2\left(x-6y\right)}{x-6y}\)
\(=x^2\)
\(2\)/
\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\dfrac{\left(x-y+z^{ }\right)^2}{\left(x-y\right)^2-z^2}\)
\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}\)
\(=\dfrac{x-y+z}{x-y-z}\)
3/
\(\dfrac{\left(n+1\right)!}{n!\left(n+2\right)}\)
\(=\dfrac{n!\left(n+1\right)}{n!\left(n+2\right)}\)
\(=\dfrac{n+1}{n+2}\)
4/
\(\dfrac{n!}{\left(n+1\right)!-n!}\)
\(=\dfrac{n!}{n!\left(n+1\right)-n!}\)
\(=\dfrac{n!}{n!\left[\left(n+1\right)-1\right]}\)
\(=\dfrac{n!}{n!.n}\)
\(=\dfrac{1}{n}\)
5/
\(\dfrac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}\)
\(=\dfrac{\left(n+1\right)!-\left(n+1\right)!\left(n+2\right)}{\left(n+1\right)!+\left(n+1\right)!\left(n+2\right)}\)
\(=\dfrac{\left(n+1\right)!\left(-n-1\right)}{\left(n+1\right)!\left(n+3\right)}\)
\(=\dfrac{-n-1}{n+3}\)
Đẳng thức đã cho tương đương với:
\(\dfrac{x^2z+y^2z-z^3+y^2x+z^2x-x^3+z^2y+x^2y-y^3}{2yxz}=1\)
\(\Leftrightarrow x^3+y^3+z^3+2xyz-x^2y-y^2z-z^2x-xy^2-yz^2-zx^2=0\)
\(\Leftrightarrow\left(x+y-z\right)\left(y+z-x\right)\left(z+x-y\right)=0\Leftrightarrow z+x=y\) (Do x + y khác z và y + z khác x).
Từ đó P = 2y (Biểu thức của P phụ thuộc vào biến y).
Áp dụng Bất đẳng thức: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (Tự chứng minh)
\(\Rightarrow C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
\(C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)