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

\(\hept{\begin{cases}x+3y-5=xy\\\frac{1}{x-1}+\frac{1}{y-2}=2\end{cases}}\)

\(\hept{\begin{cases}3y-2=xy\\\frac{1}{x-1}+\frac{1}{y-2}=2\end{cases}}\)

\(\hept{\begin{cases}\frac{3y-2}{y}=x\\\frac{1}{x-1}+\frac{1}{y-2}=2\end{cases}}\)

Ta thay \(\frac{3y-2}{y}\)vào biểu thức \(\frac{1}{x-1}+\frac{1}{y-2}\)ta đc

\(\frac{1}{\frac{3y-2}{y}-1}+\frac{1}{y-2}=2\)

\(y-3+\frac{3y-2}{y}=4y-\frac{4\left(3y-2\right)}{y}\)

\(y^2-2=4y^2-12y+8\)

\(y^2-2-4y^2+12y-8=0\)

\(-3y^2-10+12y=0\)

\(y=\orbr{\begin{cases}\frac{6-\sqrt{6}}{3}\\\frac{6+\sqrt{6}}{3}\end{cases}}\)

Tự thay y vào mà tính x , mà nếu như AD giải hpt ở lp 8 thì ta cho lak vô nghiệm đều đc ( vì số vô tỉ => vô nghiệm nha ) 

pt 1) x=y=z  Cosi 3 số 

Ta có hệ \(\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}=\frac{5}{2}\left(2\right)\end{cases}}\)

ĐK: \(x\ne0,y\ne0\)

Từ phương trình (2) ta có \(\frac{x^2y^2+1}{xy}=\frac{5}{2}\Rightarrow2x^2y^2-5xy+2=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{y}\\x=\frac{1}{2y}\end{cases}}\)

TH1: \(x=\frac{2}{y},\) thế vào phương trình (1) ta có: 

 \(\frac{2}{y}+y+\frac{y}{2}+\frac{1}{y}=\frac{9}{2}\Rightarrow\frac{3y}{2}+\frac{3}{y}=\frac{9}{2}\Rightarrow\frac{y}{2}+\frac{1}{y}=\frac{3}{2}\)

\(\Rightarrow\frac{y^2+2}{2y}=\frac{3}{2}\Rightarrow2y^2-6y+4=0\Rightarrow\orbr{\begin{cases}y=2\\y=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\) 

TH2: \(x=\frac{1}{2y},\)

Thế vào phương trình (1) ta có: 

 \(\frac{1}{2y}+y+2y+\frac{1}{y}=\frac{9}{2}\Rightarrow3y+\frac{3}{2y}=\frac{9}{2}\Rightarrow y+\frac{1}{2y}=\frac{3}{2}\)

\(\Rightarrow\frac{2y^2+1}{2y}=\frac{3}{2}\Rightarrow4y^2-6y+2=0\Rightarrow\orbr{\begin{cases}y=1\\y=\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\) (Vô nghiệm)

Tóm lại, ta có 4 cặp nghiệm \(\left(1;2\right),\left(2;1\right),\left(1;\frac{1}{2}\right),\left(\frac{1}{2};1\right)\)