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a) \(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}11\left(x+1\right)=22\\3\left(x+1\right)+2\left(x+2y\right)=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\4y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)
b) ĐK : y khác 0
\(\hept{\begin{cases}x+\frac{1}{y}=-\frac{1}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}3x+\frac{3}{y}=-\frac{3}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=-5\\3x+\frac{3}{y}=-\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-1\\-3+\frac{3}{y}=-\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\\frac{3}{y}=\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\left(tm\right)\end{cases}}\)
ĐKXĐ \(\hept{\begin{cases}x\ne2\\y\ne1\end{cases}}\)
Đặt \(\hept{\begin{cases}\frac{1}{x-2}=a\\\frac{1}{y-1}=b\end{cases}\left(a;b\ne0\right)}\)
Hệ trở thành \(\hept{\begin{cases}a+b=2\\2a-3b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a+2b=4\\2a-3b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}5b=3\\a+b=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=\frac{3}{5}\\a=\frac{7}{5}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x-2}=\frac{7}{5}\\\frac{1}{y-1}=\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2=\frac{5}{7}\\y-1=\frac{5}{3}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{19}{7}\\y=\frac{8}{3}\end{cases}}\left(TmDKXD\right)\)
a) Đặt \(a=\frac{1}{\sqrt{x-4}},b=\frac{1}{y+2}\) từ đây ta có
\(\Rightarrow\left\{{}\begin{matrix}3a+4b=7\\5a-1b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a+4b=7\\20a-4b=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}23a=23\\3a+4b=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\).
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{\sqrt{x-4}}=1\\\frac{1}{y+2}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-4=1\\y+2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)
b) Theo đề bài ta có hệ pt
\(\left\{{}\begin{matrix}u^2+v^2=65\\uv=-28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(u+v\right)^2-uv=65\\uv=-28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=65+2.\left(-28\right)=9\\uv=-28\end{matrix}\right.\)
TH1 : \(\left\{{}\begin{matrix}u+v=3\\uv=-28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=3-v\\\left(3-v\right)v=-28\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}v=-4\Rightarrow u=7\\v=7\Rightarrow u=-4\end{matrix}\right.\)
TH2 \(\left\{{}\begin{matrix}u+v=-3\\uv=-28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=-3-v\\\left(-3-v\right)v=-28\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}v=-7\Rightarrow u=4\\v=4\Rightarrow u=-7\end{matrix}\right.\)
Vậy .......