\(x^2+y^2+z^2=1\Rightarrow x^2,y^2,z^2\le1\Rightarrow-1\le x,y,z\le1\)

Ta có:\(x^3+y^3+z^3-x^2-y^2-z^2=0\)

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

Vì \(x-1\le0,y-1\le0,z-1\le0\)

\(\Rightarrow x^2\left(x-1\right)\text{​​}\le0,y^2\left(y-1\right)\le0,z^2\left(z-1\right)\le0\)

\(\Rightarrow x^2\left(x-1\right)\text{​​}+y^2\left(y-1\right)+z^2\left(z-1\right)\le0\)

Dấu "=" xảy ra khi\(\left\{{}\begin{matrix}x^2\left(x-1\right)=0\\y^2\left(y-1\right)=0\\z^2\left(z-1\right)=0\end{matrix}\right.\)

\(\Rightarrow\left(x,y,z\right)\) là bộ (0,0,1) và các hoán vị

\(\Rightarrow x^{2021}+y^{2021}+z^{2021}=1\)

Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2\)

        \(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zy\right)=x^2+y^2+z^2\)

        \(\Rightarrow2\left(xy+yz+zx\right)=0\)

        \(\Rightarrow xy+yz+zx=0\)

        \(\Rightarrow\frac{xy}{xyz}+\frac{yz}{xyz}+\frac{zx}{xyz}=0\)( Chia 2 vế cho xyz )

        \(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

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

Ta lại có : \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^3-\left(\frac{3}{x^2y}+\frac{3}{xy^2}\right)+\frac{1}{z^3}\)

               \(=\left(-\frac{1}{z}\right)^3-\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{z^3}\)

                \(=-\frac{3}{xy}\cdot-\frac{1}{z}\)\(=\frac{3}{xyz}\)

                 \(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)         ( đpcm )

\(\left(x+y+z\right)^2=x^2+y^2+z^2\)

\(\Leftrightarrow xy+yz+zx=0\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

Ta lại co:

\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}-\frac{3}{xyz}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right)=0\)

\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Ta có:

\(\left(x+y+z\right)^2=x^2+y^2+z^2\)

\(\Leftrightarrow xy+yz+zx=0\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

Ta lại có:

\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}-\frac{3}{xyz}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right)=0\)

\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

x²+y²+z²=1 => x;y;z \(\le1\)(1)

1= (x+y+z)²= x²+y²+z²+ 2(xy+yz+zx) = 1+ 2(xy+yz+zx) => xy+yz+zx=0 => xy= z(-y-x) = z(z-1)

x³+y³ =1 <=> (x+y)(x²+y² -xy)=1 <=> (1-z)(1-z²-z(z-1))=1 <=> (z-1)(2z2-z-1)= 2z³ -3z² =0 <=> z=0 hoặc z= \(\frac{3}{2}\)(loại vì lớn hơn 1)

 z=0 => x+y=1; xy= 0;

y=y(x+y) = xy+ y² = y²

=> x+y² +z³ = x+ y+ 0 = 1 (điều phải chứng minh)

Ko sai bạn ey

{ x + y + z = 1 (1)

{ x² + y² + z² = 1 (2)

{ x³ + y³ + z³ = 1 (3)

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx) 

⇒ 2(xy + yz + zx) = (x + y + z)² - (x² + y² + z²) = 1² - 1 = 0 ⇒ xy + yz + zx = 0

(x + y + z)³ = x³ + y³ + z³ + 3(x + y)(y + z)(z + x) 

⇒ 3(x + y)(y + z)(z + x) = (x + y + z)³ - (x³ + y³ + z³) = 1³ - 1 = 0

⇒ x + y = 0 hoặc y + z = 0 hoặc z + x = 0

@ Nếu  x + y = 0 ⇔ x = - y thay vào (1) ⇒ z = 1 , thay vào (2) ⇒ 2x² + 1 = 1 ⇒ x = 0; y = 0

⇒ S = 1

Tương tự cho trường hợp y + z = 0 và z + x = 0

Giải cách lớp 7 được ko bạn ????