\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)

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

Câu 1 :

Ta có :

\(\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)

\(=\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)}\)

\(=\dfrac{x^4\left(x^2+1\right)+\left(x^2+1\right)}{\left(x-1\right)}\)

\(=\dfrac{\left(x^2+1\right)\left(x^4+1\right)}{x-1}\)

Câu 2 : Ta có VT :

\(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)

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

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

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

\(=\dfrac{1}{x-y}=VP\left(đpcm\right)\)

Câu 3 :

\(\)Ta có :

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

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

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

\(\)

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).

Vậy từ giả thiết đó bạn có thể CMR P=0 đc k

Giúp mk ba mk đg cần gấp

\(C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)

Dấu "=" xảy ra khi \(x=y=z=1\)

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 \(\Leftrightarrow x=y=z=1\)

\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)

\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Dấu \("="\Leftrightarrow x=y=z=1\)

\(\sqrt{5x^2+2xy+2y^2}=\sqrt{4x^2+2xy+y^2+x^2+y^2}\ge\sqrt{4x^2+2xy+y^2+2xy}=2x+y\)

\(\Rightarrow\dfrac{1}{\sqrt{5x^2+2xy+2y^2}}\le\dfrac{1}{2x+y}=\dfrac{1}{x+x+y}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{9}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{5y^2+2yz+2z^2}}\le\dfrac{1}{9}\left(\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{\sqrt{5z^2+2zx+2x^2}}\le\dfrac{1}{9}\left(\dfrac{2}{z}+\dfrac{1}{x}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=1\)

\(P_{max}=1\) khi \(x=y=z=1\)