Giải hệ pt:
a)\(\left\{{}\begin{matrix}x^2+y^2+x+y=18\\x\left(x+1\right).y\left(y+1\right)=72\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\3y-1=xy\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}2x+3y=xy+5\\\frac{1}{x}+\frac{1}{y+1}=1\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\sqrt{\frac{x}{y}}-3\sqrt{\frac{y}{x}}=2\\x-y+xy=1\end{matrix}\right.\) e)\(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
HELP ME :((
a/ \(\left\{{}\begin{matrix}\left(x^2+x\right)+\left(y^2+y\right)=18\\\left(x^2+x\right)\left(y^2+y\right)=72\end{matrix}\right.\)
Theo Viet đảo, \(x^2+x\) và \(y^2+y\) là nghiệm của:
\(t^2-18t+72=0\Rightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=12\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\left\{2;-3\right\}\\y=\left\{3;-4\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{3;-4\right\}\\y=\left\{2;-3\right\}\end{matrix}\right.\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\x=\frac{3y-1}{y}\end{matrix}\right.\)
Nhận thấy \(y=\frac{1}{3}\) không phải nghiệm
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\\frac{1}{x}=\frac{y}{3y-1}\end{matrix}\right.\) \(\Rightarrow\frac{y}{3y-1}+\frac{1}{y+1}=1\)
\(\Leftrightarrow y\left(y+1\right)+3y-1=\left(3y-1\right)\left(y+1\right)\)
\(\Leftrightarrow y^2-y=0\Rightarrow\left[{}\begin{matrix}y=0\left(l\right)\\y=1\end{matrix}\right.\) \(\Rightarrow x=2\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=xy+5\\y+1=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y-1=5\\y+1=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\y+1=xy\end{matrix}\right.\)
\(\Rightarrow y+1=\left(3-y\right)y\)
\(\Leftrightarrow y^2-2y+1=0\Rightarrow y=1\Rightarrow x=2\)
d/ ĐKXĐ: ...
Xét pt trên, đặt \(\sqrt{\frac{x}{y}}=t>0\)
\(\Rightarrow t-\frac{3}{t}=2\Leftrightarrow t^2-2t-3=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=3\end{matrix}\right.\)
\(\Rightarrow\sqrt{\frac{x}{y}}=3\Rightarrow x=9y\)
Thế xuống pt dưới:
\(9y-y+9y^2=1\)
\(\Leftrightarrow9y^2+8y-1=0\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=-9\\y=\frac{1}{9}\Rightarrow x=1\end{matrix}\right.\)
e/ ĐKXĐ: ...
Biến đổi pt đầu:
\(\Leftrightarrow x^2-xy-2y^2-\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)-\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y-1\right)=0\)
Do \(\left\{{}\begin{matrix}x\ge1\\y\ge0\end{matrix}\right.\) \(\Rightarrow x+y>0\)
\(\Rightarrow x-2y-1=0\Rightarrow x=2y+1\)
Thay xuống dưới:
\(\left(2y+1\right)\sqrt{2y}-y\sqrt{2y}=2y+2\)
\(\Leftrightarrow\left(y+1\right)\sqrt{2y}-2\left(y+1\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(\sqrt{2y}-2\right)=0\Rightarrow\left[{}\begin{matrix}y=-1\left(l\right)\\y=2\end{matrix}\right.\) \(\Rightarrow x=5\)
