Giải hpt:
\(\left\{\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)
Những câu hỏi liên quan
Giải hpt:
\(\left\{\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)
\(\left\{\begin{matrix}x^2+x-xy-2y=0\\x^2+y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}\left(x+y+1\right)\left(2y-x\right)=0\\x^2+y^2=1\end{matrix}\right.\)
Với x + y + 1 = 0 \(\Rightarrow\)x = - y - 1 thế vô pt dưới được
\(\left(-y-1\right)^2+y^2=1\)
\(\Leftrightarrow\left[\begin{matrix}y=0\\y=-1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Với 2y - x = 0 \(\Rightarrow\)2y = x thế vào pt dưới được
\(\left(2y\right)^2+y^2=1\)
\(\Leftrightarrow\left[\begin{matrix}y=\frac{1}{\sqrt{5}}\\y=-\frac{1}{\sqrt{5}}\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=\frac{2}{\sqrt{5}}\\x=-\frac{2}{\sqrt{5}}\end{matrix}\right.\)
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giải hệ pt :
a,\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3-30=0\\x^2y+x\left(1+y+y^2\right)+y-11=0\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}xy^2-2y+3x^2=0\\y^2+x^2y+2x=0\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}3xy+2y=5\\2xy\left(x+y\right)+y^2=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)
TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)
Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)
TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)
2 câu dưới hình như em hỏi rồi?
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giải hpt:
1, \(\left\{{}\begin{matrix}x^2y^2+4=2y^2\\\left(xy+2\right)\left(y-x\right)=x^3y^3\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^2+y^2-4xy\left(\dfrac{2}{x-y}-1\right)=4\left(4+xy\right)\\\sqrt{x-y}+3\sqrt{2y^2-y+1}=2y^2-x+3\end{matrix}\right.\)
Giải hệ
a) \(\left\{{}\begin{matrix}xy+y^2=1+y\\x^2+2y^2+2xy=4+x\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-2y^2-xy+2y-x=0\\x^2-y^2+6xy+12=0\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}2xy+2y^2=2+2y\\x^2+2y^2+2xy=4+x\end{matrix}\right.\)
\(\Rightarrow x^2+4xy+4y^2=x+2y+6\)
\(\Leftrightarrow\left(x+2y\right)^2-\left(x+2y\right)-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2y=3\\x+2y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-2y\\x=-2-2y\end{matrix}\right.\)
Thế vào pt đầu...
b.
Từ pt đầu:
\(\left(x^2-xy-2y^2\right)-\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)-\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1-y\\x=2y\end{matrix}\right.\)
Thế xuống pt dưới...
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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: \(\left\{{}\begin{matrix}x^2+2y^2=3\\x+y^2+xy=1\end{matrix}\right.\)
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 các hệ phương trình sau : a, left{{}begin{matrix}x^2+xyy^2+13x+yy^2+3end{matrix}right.b,left{{}begin{matrix}x^2-y^24x-2y-3x^2+y^25end{matrix}right.c, left{{}begin{matrix}x^2+x-xy-2y^2-2y0x^2+y^21end{matrix}right.d,left{{}begin{matrix}2left(y+zright)yzxy+yz+zx108xyz180end{matrix}right.
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Giải các hệ phương trình sau :
a, \(\left\{{}\begin{matrix}x^2+xy=y^2+1\\3x+y=y^2+3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x^2-y^2=4x-2y-3\\x^2+y^2=5\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}2\left(y+z\right)=yz\\xy+yz+zx=108\\xyz=180\end{matrix}\right.\)
Giải HPT
\(\left\{{}\begin{matrix}xy^2+3x^2=2y\\x^2y+y^2=-2x\end{matrix}\right.\)