a) \(\hept{\begin{cases}\left(x-1\right)\left(2x+y\right)=0\\\left(y+1\right)\left(2y-x\right)=0\end{cases}}\)
\(\cdot x=1\Rightarrow\hept{\begin{cases}0=0\\\left(y+1\right)\left(2y-1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\y=-1;y=\frac{1}{2}\end{cases}}\)
\(\cdot y=-1\Rightarrow\hept{\begin{cases}\left(x-1\right)\left(2x-1\right)=0\\0=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1;x=\frac{1}{2}\\0=0\end{cases}}\)
\(\cdot x=2y\Rightarrow\hept{\begin{cases}\left(2y-1\right)5y=0\\0=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=0\\y=\frac{1}{2}\Rightarrow x=1\end{cases}}\)
\(y=-2x\Rightarrow\hept{\begin{cases}0=0\\\left(1-2x\right)5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\Rightarrow y=-1\\x=0\Rightarrow y=0\end{cases}}\)

b) \(\hept{\begin{cases}x+y=\frac{21}{8}\\\frac{x}{y}+\frac{y}{x}=\frac{37}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\\left(\frac{21}{8}-y\right)^2+y^2=\frac{37}{6}y\left(\frac{21}{8}-y\right)\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\2y^2-\frac{21}{4}y+\frac{441}{64}=-\frac{37}{6}y^2+\frac{259}{16}y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\1568y^2-4116y+1323=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{8}\\y=\frac{9}{4}\end{cases}}hay\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{3}{8}\end{cases}}\)

c) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x-y\right)^2=-4x^2y^2+2xy\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}8x^2y^2-4x^2y-4xy^2+x^2+y^2-2xy+2xy=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4x^2y^2-4x^2y+x^2+4x^2y^2-4xy^2+y^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x\right)^2+\left(2xy-y\right)^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=\frac{-1}{2}\end{cases}}\)
d) \(\hept{\begin{cases}xy+x+y=71\\x^2y+xy^2=880\end{cases}}\). Đặt \(\hept{\begin{cases}x+y=S\\xy=P\end{cases}}\), ta có: \(\hept{\begin{cases}S+P=71\\SP=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P\left(71-P\right)=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P^2-71P+880=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=16\\P=55\end{cases}}hay\hept{\begin{cases}S=55\\P=16\end{cases}}\)
\(\cdot\hept{\begin{cases}S=16\\P=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=16\\xy=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y\left(16-y\right)=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y^2-16y+55=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=11\end{cases}}hay\hept{\begin{cases}x=11\\y=5\end{cases}}\)

\(\cdot\hept{\begin{cases}S=55\\P=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=55\\xy=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y\left(55-y\right)=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y^2-55y+16=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{55-3\sqrt{329}}{2}\\y=\frac{55+3\sqrt{329}}{2}\end{cases}}hay\hept{\begin{cases}x=\frac{55+3\sqrt{329}}{2}\\y=\frac{55-3\sqrt{329}}{2}\end{cases}}\)

e) \(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{cases}}\). Đặt \(\hept{\begin{cases}S=\sqrt{x}+\sqrt{y}\\P=\sqrt{xy}\end{cases}}\), ta có \(\hept{\begin{cases}SP=12\\P\left(S^2-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\P\left(\frac{144}{P^2}-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\2P^4+28P^2-144P=0\end{cases}}\)
Tự làm tiếp nhá! Đuối lắm luôn

a/ Đảo ngược lại rồi đặc \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)

b/ Dễ thấy vai trò x, y, z như nhau nên ta chỉ cần xét 1 trường hợp tiêu biểu thôi.

Xét \(x>y>z\)

\(\Rightarrow\frac{1}{x}< \frac{1}{y}< \frac{1}{z}\)

\(\Rightarrow x+\frac{1}{y}>z+\frac{1}{x}\)(trái giả thuyết)

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

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

\(\Leftrightarrow x=1\)

sai lớp :>>>

Rõ ràng \(x=y=z=0\)   là nghiệm của hệ

Với \(xyz\ne0\), Ta có

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

\(z=\frac{3y^3}{y^4+y^2+1}\le\frac{3y^3}{3y^2}=y\)

\(x=\frac{4z^4}{z^6+z^4+z^2+1}\le\frac{4z^4}{4z^3}=z\)

Suy ra \(y\le x\le z\le y\Rightarrow x=y=z\)

Từ pt thứ nhất của hệ suy ra 

\(\frac{2x^2}{x^2+1}=x\Leftrightarrow2x=1=x^2\)( vì \(x\ne0\))\(\Leftrightarrow x=1\)

Vậy hệ pt có hai nghiệm \(\left(0,0,0\right)\)và \(\left(1,1,1\right)\)

Khong mat tinh tong quat gia su \(x\ge y\ge z\)

Ta co:

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

\(z=\frac{3y^3}{1+y^2+y^4}\le\frac{3y^3}{3y^2}=y\)

\(\Rightarrow x\ge y\ge z\) (đúng)

Dau'=' xay ra khi \(x=y=z=1\)

1 slot tối làm cho.Giờ đi học đã =))

\(\hept{\begin{cases}\frac{x}{3}+\frac{y}{12}-\frac{z}{4}=1\\\frac{x}{10}+\frac{y}{5}+\frac{z}{3}=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x+y-3z=12\\3x+6y+10z=30\end{cases}}\)

\(\Rightarrow7\left(x+y+z\right)=42\)

\(\Leftrightarrow x+y+z=6\)

\(\hept{\begin{cases}\frac{x}{3}+\frac{y}{12}-\frac{z}{4}=1\\\frac{x}{10}+\frac{y}{5}+\frac{z}{3}=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x+y-3z=12\\3x+6y+10z=30\end{cases}}\)

=> 7 ( x + y +z ) = 42 

<=> x + y + z = 6