Giải hệ phương trình: \(\hept{\begin{cases}\left(x^2-y^2\right)^2+3\left(xy+1\right)^2=5\\x+\frac{y^2}{x+y}=2\end{cases}}\)
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Đặt \(\hept{\begin{cases}x+y=a\\x-y=b\end{cases}}\)\(\Rightarrow\)ta có hệ \(\hept{\begin{cases}2a+3b=4\\a+2b=5\end{cases}}\Rightarrow\hept{\begin{cases}a=-7\\b=6\end{cases}}\)Từ đó ta có \(\hept{\begin{cases}x+y=-7\\x-y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{13}{2}\end{cases}}\)PS: Cái đề chỗ 3(x+y) phải thành 3(x-y) chứ
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: xy\(\ne\)0
HPT đã cho tương đương: \(\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=9\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=9\end{cases}}\)
Đặt \(\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=S\\\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=P\end{cases}}\)
Hệ trở thành:
\(\hept{\begin{cases}S^2-2P=9\\S=5\end{cases}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{x}=2;y+\frac{1}{y}=3\\x+\frac{1}{x}=3;y+\frac{1}{y}=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1;y=\frac{3\pm\sqrt{5}}{2}\\x=\frac{3\pm\sqrt{5}}{2};y=1\end{cases}}}\)
Vậy HPT đã cho có nghiệm (x;y)=\(\left(1;\frac{3\pm\sqrt{5}}{2}\right);\left(\frac{3\pm\sqrt{5}}{2};1\right)\)
\(\hept{\begin{cases}\left(x+y\right)\left(1+\frac{1}{xy}\right)=5\\\left(x^2+y^2\right)\left(1+\frac{1}{x^2y^2}\right)=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{x}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=13\end{cases}}\)
\(\left(x+\frac{1}{x};y+\frac{1}{y}\right)\rightarrow\left(a;b\right)\)
Hệ pt \(\Leftrightarrow\hept{\begin{cases}a+b=5\\a^2+b^2=13\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=5\\\left(a+b\right)^2-2ab=13\end{cases}\Leftrightarrow\hept{\begin{cases}a+b=5\\ab=6\end{cases}}}\)
Tự làm nốt nhé
Đặt \(\hept{\begin{cases}x+y=a\\xy=b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(x^2-y^2\right)^2+3\left(xy+1\right)^2=5\\x+\frac{y^2}{x+y}=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2a-b\right)^2-b^2+6b=2\\a^2-b-2a=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=\frac{2a^2-1}{2a-3}\\b=\frac{2a^2-4a}{2}\end{cases}}\)
\(\Rightarrow\frac{2a^2-1}{2a-3}=\frac{2a^2-4a}{2}=\frac{4a-1}{2a-5}\)
\(\Leftrightarrow\left(a-1\right)\left(a^2-7a-1\right)=0\)
Làm nốt