giải hpt: \(\left\{{}\begin{matrix}xy+x+y=0\\xy^2-4=x^2\end{matrix}\right.\)
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giải hpt:
1,\(\left\{{}\begin{matrix}x^2y^2-2x+y^2=0\\2x^2-4x+3+y^3=0\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\left(x^2-xy\right)\left(xy-y^2\right)=25\\\sqrt{x^2-xy}+\sqrt{xy-y^2}=3\left(x-y\right)\end{matrix}\right.\)
Giải hpt \(\left\{{}\begin{matrix}x^2+x^3y-xy^2+xy-y=1\\x^4+y^2-xy\left(2x-1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(xy+1\right)-y\left(xy+1\right)+xy+1=2\\\left(x^2-y\right)^2+xy+1=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y+1\right)\left(xy+1\right)=2\\\left(x^2-y\right)^2+xy+1=2\end{matrix}\right.\)
\(\Rightarrow\left(x^2-y+1\right)\left(xy+1\right)-\left(x^2-y\right)^2-\left(xy+1\right)=0\)
\(\Leftrightarrow\left(xy+1\right)\left(x^2-y\right)-\left(x^2-y\right)^2=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(xy+1-x^2+y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=x^2\\xy+1=x^2-y\end{matrix}\right.\) thay xuống pt dưới:
- Với \(y=x^2\) thay xuống pt dưới \(\Rightarrow x^3=1\)
- Với \(xy+1=x^2-y\) thay xuống dưới:
\(\left\{{}\begin{matrix}xy+1=x^2-y\\2\left(xy+1\right)=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}xy+1=x^2-y\\xy=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0;y=-1\\y=0;x^2=1\end{matrix}\right.\)
Giải HPT \(\left\{{}\begin{matrix}x^2+xy+y^2=4\\x+xy+y=2\end{matrix}\right.\)
Giải HPT:\(\left\{{}\begin{matrix}4x^2+y^2+4xy=4\\x^2+y^2-2\left(xy-8\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y\right)^2=4\\\left(x-y\right)^2=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x+y=\pm2\\x-y=\pm4\end{matrix}\right.\)
Giải hệ phương trình
left{{}begin{matrix}x^2+y^2+xy13x^4+y^4+x^2y^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^2-xy13left(x^2+y^2right)^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^213+xyleft[left(x+yright)^2-2xyright]^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^2-xy13left(13-xyright)^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}xy3left(x+yright)^216end{matri...
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Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?
Giải HPT
\(\left\{{}\begin{matrix}2x^2-xy+3y^2=7x+12y-1\\x-y+1=0\end{matrix}\right.\)
\(PT\left(2\right)\Leftrightarrow x=y-1\\ PT\left(1\right)\Leftrightarrow2\left(y-1\right)^2+y\left(1-y\right)+3y^2=7\left(y-1\right)+12y-1\\ \Leftrightarrow2y^2-11y+5=0\\ \Leftrightarrow\left[{}\begin{matrix}y=5\Leftrightarrow x=4\\y=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
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giải hpt: \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left(y+\sqrt{xy}+x-x^2\right)=4\end{matrix}\right.\)
Giải PT và HPT:
1)left{{}begin{matrix}xy+x+y3frac{1}{x^2+2x}+frac{1}{y^2+2y}frac{2}{3}end{matrix}right.
2)left(sqrt{x+4}-2right)left(sqrt{4-x}+2right)2x
3)left{{}begin{matrix}xyleft(x+yright)29xyleft(3x-yright)+626x^3-2y^3end{matrix}right.
4)left{{}begin{matrix}x^2-2xy+x-2y+30y^2-x^2+2xy+2x-20end{matrix}right.
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Giải PT và HPT:
1)\(\left\{{}\begin{matrix}xy+x+y=3\\\frac{1}{x^2+2x}+\frac{1}{y^2+2y}=\frac{2}{3}\end{matrix}\right.\)
2)\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
3)\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\9xy\left(3x-y\right)+6=26x^3-2y^3\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^2-2xy+x-2y+3=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)
Giải HPT: \(\left\{{}\begin{matrix}xy+y^2+x-5y=0\\\left(x+y\right)\dfrac{x}{y}=6\end{matrix}\right.\)
Đk: \(y\ne0\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy-6y=xy+y^2+x-5y\\x+y=\dfrac{6y}{x}\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2-y-x=0\\x+y=\dfrac{6y}{x}\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(x-y\right)-\left(x+y\right)=0\\x+y=\dfrac{6y}{x}\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6y}{x}\left(x-y-1\right)=0\\x+y=\dfrac{6y}{x}\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1+y\\x+y=\dfrac{6y}{x}\\x,y\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1+y\\1+2y=\dfrac{6y}{1+y}\\x,y\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1+y\\1+2y+y+2y^2=6y\\x,y\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1+y\\2y^2-3y+1=0\left(@\right)\\x,y\ne0\end{matrix}\right.\)
(@) \(\Leftrightarrow\left[{}\begin{matrix}y=1\left(N\right)\\y=\dfrac{1}{2}\left(N\right)\end{matrix}\right.\)
Với y=1, ta có x=2 (N)
Với y= 1/2 , ta có x= 3/2 (N)
KL : nếu x= 2 thì y=1
nếu x=3/2 thì y=1/2
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