Sử dụng bất đẳng thức: 

\(x^3+y^3\ge3xy\left(x+y\right)\)

Có: \(M=2018\left(\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\right)\)

\(M\le2018\left(\frac{xyz}{xy\left(x+y\right)+xyz}+\frac{xyz}{yz\left(y+z\right)+xyz}+\frac{xyz}{xz\left(x+z\right)+xyz}\right)\)

\(M\le2018\left(\frac{xyz}{xy\left(x+y+z\right)}+\frac{xyz}{yz\left(x+y+z\right)}+\frac{xyz}{xz\left(x+y+z\right)}\right)\)

\(M\le2018\left(\frac{x+y+z}{x+y+z}\right)=2018\)

Vậy Max M=2018 khi x=y=z=1

Sửa lại \(x^3+y^3\ge xy\left(x+y\right)\)

Xin lỗi

Bài 1 : 

Ta có : 

\(x^7+\frac{1}{x^7}=\left(x^3+\frac{1}{x^3}\right)\left(x^4+\frac{1}{x^4}\right)-\left(x+\frac{1}{x}\right)\)

\(\left(x+\frac{1}{x}\right)=a\Leftrightarrow\left(x+\frac{1}{x}\right)^2=a^2\)

\(\Leftrightarrow x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=a^2\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-x.\frac{1}{x}+\frac{1}{x^2}\right)\)

               \(=a\left(x^2+\frac{1}{x^2}-1\right)=a\left(a^2-3\right)\)

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

                   \(=\left(a^2-2\right)^2-2=a^4-4a^2+4-2\)

                                                               \(=a^4-4a^2+2\)

\(\Rightarrow x^7+\frac{1}{x^7}=a.\left(a^2-3\right).\left(a^4-4a^2+2\right)-a\)

                      \(=\left(a^3-3a\right)\left(a^4-4a^2+2\right)-a\)

                         \(=a^7-4a^5+2a^3-3a^5+12a^3-6a-a\)

                          \(=a^7-7a^5+14a^3-7a\)

Bài 2 : 

Ta có : 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=\frac{2}{xy}-\frac{1}{z^2}\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}+\frac{2}{yz}+\frac{2}{zx}=0\)

\(\Rightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)

\(\Rightarrow\frac{1}{x}+\frac{1}{z}=\frac{1}{y}+\frac{1}{z}=0\) vì \(\left(\frac{1}{x}+\frac{1}{z}\right)^2,\left(\frac{1}{y}+\frac{1}{z}\right)^2\ge0\)

\(\Rightarrow x=y=-z\)

\(\Rightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\Rightarrow-\frac{1}{z}=2\Rightarrow z=-\frac{1}{2}\)

\(\Rightarrow x=y=\frac{1}{2}\)

\(\Rightarrow x+2y+z=\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}=1\)

\(\Rightarrow P=1\)

 Ko biết Anh gì ơi

\(2x^2+2y^2+z^2-2x+2y+2xy+2yz+2zx+2=0\)

\(\Leftrightarrow\)\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+y\right)^2+\left(y+z\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

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

\(\Rightarrow\)\(A=x^{2018}+y^{2018}+z^{2018}=1^{2018}+\left(-1\right)^{2018}+1^{2018}=3\)

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