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\(\hept{\begin{cases}x^2+8=xy^2+2x\left(1\right)\\y^2+8=x^2y+2y\left(2\right)\end{cases}}\)
Xét: \(x^2+8=xy^2+2x\)
<=> \(x\left(y^2+2\right)=x^2+8\ge8>0\)mà \(y^2+2>0\) với mọi x; y
=> \(x>0\)tương tự \(y>0\)(3)
Xét \(x^2+8=xy^2+2x\)
<=> \(y^2+2=x+\frac{8}{x}\ge2\sqrt{8}\)<=> \(y^2\ge2\sqrt{8}-2\)
<=> \(\orbr{\begin{cases}y\ge\sqrt{2\sqrt{8}-2}\\y\le-\sqrt{2\sqrt{8}-2}\end{cases}}\)tương tự \(\orbr{\begin{cases}x\ge\sqrt{2\sqrt{8}-2}\\x\le-\sqrt{2\sqrt{8}-2}\end{cases}}\)(4)
Từ (3) và (4) => \(x;y\ge\sqrt{2\sqrt{8}-2}\)(@@)
Lấy (1) - ( 2) ta có: \(x^2-y^2=xy^2-x^2y+2x-2y\)
<=> \(\left(x-y\right)\left(x+y\right)+xy\left(x-y\right)-2\left(x-y\right)=0\)
<=> \(\left(x-y\right)\left(x+y+xy-2\right)=0\)(5)
Với \(x;y\ge\sqrt{2\sqrt{8}-2}\) ta có: \(x+y+xy-2>0\)
Do đó: (5) <=> x = y
Thế vào (1) ta có: \(x^3-x^2+2x-8=0\Leftrightarrow x=2\)thỏa mãn (@@)
Vậy:...
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\(\hept{\begin{cases}x^2+8=xy^2+2x\left(1\right)\\y^2+8=x^2y+2y\left(2\right)\end{cases}}\)
\(\left(1\right)-\left(2\right)\Leftrightarrow\left(x-y\right)\left(x+y\right)=-xy\left(x-y\right)+2\left(x-y\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+xy-2\right)=0\)
Đến đây dễ r :)))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\hept{\begin{cases}x^2y+xy^2+x+y=9\\xy+2x+2y=8\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}xy\left(x+y\right)+x+y=9\\xy+2x+2y=8\end{cases}}\)
Đặt \(\hept{\begin{cases}x+y=S\\xy=P\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}P.S+S=9\\P+2S=8\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}S\left(8-2S\right)+S=9\\P=8-2S\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2S^2-9S+9=0\\P=8-2S\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}S=\frac{3}{2}\\P=5\end{cases}}\)hoặc \(\Rightarrow\hept{\begin{cases}S=3\\P=2\end{cases}}\)
Thế vô giải tiếp.
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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
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1) \(\hept{\begin{cases}x^2+y^2-xy=1\\x+x^2y=2y^3\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2+y^2=1+xy\\x\left(1+xy\right)=2y^3\end{cases}\Rightarrow x\left(x^2+y^2\right)=2y^3}\)
\(\Leftrightarrow\left(x^3-y^3\right)+\left(xy^2-y^3\right)=0\Leftrightarrow\left(x-y\right)\left(x^2+y^2+xy\right)+y^2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+2y^2+xy\right)=0\Leftrightarrow\orbr{\begin{cases}x=y\\x^2+2y^2+xy=0\end{cases}}\)
+) \(x=y\Rightarrow\hept{\begin{cases}y^2+y^2-y^2=1\\y+y^3=2y^3\end{cases}\Rightarrow}x=y=\pm1\)
+) \(x^2+2y^2+xy=0\)Vì y=0 không là nghiệm của hệ nên ta chia 2 vế phương trình cho y2:
\(\Rightarrow\left(\frac{x}{y}\right)^2+\frac{x}{y}+2=0\)( Vô nghiệm)
Vậy hệ có nghiệm (1;1),(-1;-1).
2/ \(\hept{\begin{cases}x+y=\sqrt{x+3y}\\x^2+y^2+xy=3\end{cases}\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=x+3y\\x^2+y^2+xy=3\end{cases}}}\Rightarrow xy=x+3y-3\)
\(\Leftrightarrow\left(x-xy\right)+\left(3y-3\right)\Leftrightarrow\left(x-3\right)\left(1-y\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\Rightarrow y\in\varnothing\\y=1\Rightarrow x=1\end{cases}}\)
Vậy hệ có nghiệm (1;1).
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Ta có: \(8-y^2=\left|xy-4\right|\ge0\Rightarrow y^2\le8\) (1)
\(x^2+2=xy\Rightarrow x^2-xy+2=0\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2-\dfrac{y^2}{4}+2=0\Leftrightarrow\dfrac{y^2}{4}-2=\left(x-\dfrac{y}{2}\right)^2\ge0\)
\(\Rightarrow y^2\ge8\) (2)
Từ (1); (2) \(\Rightarrow y^2=8\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}y^2=8\\xy-4=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\) \(\Leftrightarrow...\)
\(pt\left(2\right)\Leftrightarrow\left(x^4\right)^2+\left(y^4\right)^2=35\)
\(\Leftrightarrow\left(x^4+y^4\right)^2-2x^4y^4=35\)
\(\Leftrightarrow\left(x^4+y^4\right)^2-2x^4y^4=35\)
\(\Leftrightarrow\left(\left(x^2\right)^2+\left(y^2\right)^2\right)^2-2x^4y^4=35\)
\(\Leftrightarrow\left(\left(x^2+y^2\right)^2-2x^2y^2\right)^2-2\left(xy\right)^4=35\)
\(\Leftrightarrow\left[\left(\left(x+y\right)^2-2xy\right)^2-2\left(xy\right)^2\right]^2-2\left(xy\right)^4=35\)
Và \(pt\left(1\right)\Leftrightarrow xy\left(x+y\right)=30\)
Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\Rightarrow S^2\ge4P\) thì có:
\(\hept{\begin{cases}\left[\left(S^2-2P\right)^2-2P^2\right]^2-2P^4=35\\SP=30\end{cases}}\)
Thay lẫn lộn vào nhau giải ra thì có....
Thắng Nguyễn cách này không khả thi đâu. You cứ giải đến cuối sẽ thấy.