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Đặt \(x+1=u;y-2=v\)
Hệ trở thành \(\hept{\begin{cases}\frac{2}{u}+\frac{1}{v}=\frac{1}{3}\\\frac{3}{u}+\frac{2}{v}=\frac{1}{5}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{u}+\frac{2}{v}=\frac{2}{3}\left(1\right)\\\frac{3}{u}+\frac{2}{v}=\frac{1}{5}\left(2\right)\end{cases}}\)
Lấy (1) - (2), ta được\(\frac{1}{u}=\frac{7}{15}\Leftrightarrow u=\frac{15}{7}\)
\(\Rightarrow x=\frac{15}{7}-1=\frac{8}{7}\)
Từ đó tính được \(y=\frac{1}{3}\)
Vậy hệ có 1 nghiệm \(\left(\frac{8}{7};\frac{1}{3}\right)\)
<=> \(\hept{\begin{cases}\frac{4}{x+1}+\frac{2}{y-2}=\frac{2}{3}\\\frac{3}{x+1}+\frac{2}{y-2}=\frac{1}{5}\end{cases}}\)
<=> \(\hept{\begin{cases}\frac{1}{x+1}=\frac{7}{15}\\\frac{3}{x+1}+\frac{2}{y-2}=\frac{1}{5}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{8}{7}\\y=\frac{7}{5}\end{cases}}\)
\(\hept{\begin{cases}\frac{2}{x+y}+\frac{1}{x-y}=3\\\frac{1}{x+y}-\frac{3}{x-y}=1\end{cases}}\)
Đặt: \(u=\frac{1}{x+y};v=\frac{1}{x-y}\). Ta có:
\(\hept{\begin{cases}2u+v=3\\u-3v=1\end{cases}}\)
\(\hept{\begin{cases}2u+v=3\\2u-6v=2\end{cases}}\)<=> 7v=1 => \(v=\frac{1}{7};u=\frac{10}{7}\)
\(< =>\hept{\begin{cases}\frac{1}{x+y}=\frac{10}{7}\\\frac{1}{x-y}=\frac{1}{7}\end{cases}}\) <=> \(\hept{\begin{cases}10x+10y=7\\x-y=7\end{cases}}\)<=> 10(y+7)+10y=7
<=> 20y+70=7
=> \(y=-\frac{63}{20}\); \(x=\frac{77}{20}\)
a = \(\frac{1}{x+y}\)
b = \(\frac{1}{x-y}\)
=>
\(\hept{\begin{cases}2a+b=3\\a-3b=1\end{cases}}\)
<=>
\(\hept{\begin{cases}2a+b=3\\2a-6b=2\end{cases}}\)
Trừ 2 vế PT
=> 7b = 1
=> b = 1/7
=> a = 10/7
=>
\(\hept{\begin{cases}x+y=\frac{7}{10}\\x-y=7\end{cases}}\)
<=>
\(\hept{\begin{cases}x=\frac{77}{20}\\y=-\frac{63}{20}\end{cases}}\)
<=> \(\hept{\begin{cases}\frac{2}{x}+\frac{4}{y}=\frac{2}{3}\\\frac{2}{x}-\frac{3}{y}=\frac{1}{4}\end{cases}}\)
<=> \(\hept{\begin{cases}\frac{7}{y}=\frac{5}{12}\\\frac{1}{x}+\frac{2}{y}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{14}{3}\\y=\frac{84}{5}\end{cases}}\)
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
Đk: x, y khác 0
Đặt: \(\frac{1}{x}=u;\frac{1}{y}=v\)
ta có hệ phương trình:
\(\hept{\begin{cases}u-v=1\\2u+4v=5\end{cases}}\)Giải u; v sau đó tìm x, y.
