Pt đầu chắc là sai đề (chắc chắn), bạn kiểm tra lại

Với pt sau:

Nhận thấy một ẩn bằng 0 thì 2 ẩn còn lại cũng bằng 0, do đó \(\left(x;y;z\right)=\left(0;0;0\right)\) là 1 nghiệm

Với \(x;y;z\ne0\)

Từ pt đầu ta suy ra \(y>0\) , từ đó suy ra \(z>0\) từ pt 2 và hiển nhiên \(x>0\) từ pt 3

Do đó:

\(\left\{{}\begin{matrix}y=\dfrac{2x^2}{x^2+1}\le\dfrac{2x^2}{2x}=x\\z=\dfrac{3y^3}{y^4+y^2+1}\le\dfrac{3y^3}{3\sqrt[3]{y^4.y^2.1}}=y\\x=\dfrac{4z^4}{z^6+z^4+z^2+1}\le\dfrac{4z^4}{4\sqrt[4]{z^6z^4z^2}}=z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y\le x\\z\le y\\x\le z\end{matrix}\right.\) \(\Rightarrow x=y=z\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)

Vậy nghiệm của hệ là \(\left(x;y;z\right)=\left(0;0;0\right);\left(1;1;1\right)\)

Ta có \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\)

\(\Leftrightarrow\dfrac{\left(yz\right)^2+\left(xz\right)^2+\left(xy\right)^2+2xyz}{\left(xyz\right)^2}=1\)

<=> (xy)² + (yz)² + (zx)² + 2xyz = (xyz)² 

<=> (xy)² + (yz)² + (xz)² + 2xyz(x + y + z) = (xyz)² 

<=> (xy + yz + zx)² = (xyz)² 

<=> \(\left[{}\begin{matrix}xy+yz+zx=xyz\\xy+yz+zx=-xyz\end{matrix}\right.\)

+) Khi xy + yz + zx = -xyz 

=> \(\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-1< 0\left(\text{loại}\right)\)

=> xy + yz + zx = xyz 

<=> \(xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=xyz\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1\right)=0\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

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

<=> \(\left(x+y\right)\left(\dfrac{1}{xz+yz+z^2}+\dfrac{1}{xy}\right)=0\)

<=> \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(zx+yz+z^2\right)xy}=0\)

<=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

Khi x = -y => y = 1 => P = 1

Tương tự y = -z ; z = -x được P = 1

Vậy P = 1 

a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-15y+25x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-5\left(3y-5x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)

Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)

=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}x-y=2\\y-3z=2\\-3x-2y+z=-2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3\left(y+2\right)-2\left(3z+2\right)+z=-2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3y-6-6z-4+z=-2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3y-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3\left(3z+2\right)-5z=8\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-9z-6-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-14z=14\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\left(-1\right)+2=1\\y=3\left(-1\right)+2=-1\\z=-1\end{matrix}\right.\)

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