Giải hệ phương trình sau: \(\left\{{}\begin{matrix}x^2+xy+y^2+4x+3y=0\\xy+x+2y=0\end{matrix}\right.\)
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 hệ phương trình:
\(\left\{{}\begin{matrix}xy+3y^2+x=3\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}xy+3y^2+x=3\left(1\right)\\x^2+xy-2y^2\left(2\right)\end{matrix}\right.\)
\(pt\left(2\right)\Leftrightarrow\left(x^2-y^2\right)+y\left(x-y\right)=0\Leftrightarrow\left(x-y\right)\left(x+2y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)
+) Với x=y, thay vào pt (1) ta có: \(4x^2+x-3=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
=> \(x=y=-1;x=y=\dfrac{3}{4}\)
+) Với \(x=-2y\), thay vào pt(1) ta có: \(y^2-2y-3=0\Leftrightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=2\\y=3\Rightarrow x=-6\end{matrix}\right.\)
Vậy hpt có 4 nghiệm: \(\left(x;y\right)\in\left\{\left(-1;-1\right),\left(\dfrac{3}{4};\dfrac{3}{4}\right),\left(2;-1\right),\left(-6;3\right)\right\}\)
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
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?
Giải hệ phương trình \(\left\{{}\begin{matrix}x^2-xy+y-7=0\\x^2+xy-2y=4\left(x-1\right)\end{matrix}\right.\)
Biến đổi pt dưới:
\(x^2-4x+4+y\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+y\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2+y\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2-y\end{matrix}\right.\)
Thay vào pt đầu giải bt
Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2-xy+y-7=0\\x^2+xy-2y=4\left(x-1\right)\end{matrix}\right.\)
Giải hệ phương trình:\(\left\{{}\begin{matrix}x^3+xy^2+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+2y^2+xy+2x-4=0\end{matrix}\right.\)
giải hpt:
a) \(\left\{{}\begin{matrix}4x+9y=6\\3x^2+6xy-x+3y=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(x+y+2\right)\left(2x+2y-1\right)=0\\3x^2-32y^2+5=0\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}2x^2-xy+3y^2=7x+12y-1\\x-y+1=0\end{matrix}\right.\)
giải hệ pt :
a, \(\left\{{}\begin{matrix}3y=\dfrac{y^2+2}{x^2}\\3x=\dfrac{x^2+2}{y^2}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^2y+xy^2+x-5y=0\\2xy+y^2-5y+1=0\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+y^2+xy+2y+x=2\\2x^2-y^2-2y-2=0\end{matrix}\right.\)
ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html
b.
Với \(xy=0\) không là nghiệm
Với \(xy\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y^2+1\right)=y\left(5-x^2\right)\\y^2+1=y\left(5-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y^2+1}{y}=\dfrac{5-x^2}{x}\\\dfrac{y^2+1}{y}=5-2x\end{matrix}\right.\)
\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)
\(\Leftrightarrow5-x^2=5x-2x^2\)
\(\Leftrightarrow...\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x\left(y+1\right)+\left(y+1\right)^2=3\\2x^2-\left(y+1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x\left(y+1\right)+\left(y+1\right)^2=3\\6x^2-3\left(y+1\right)^2=3\end{matrix}\right.\)
\(\Rightarrow5x^2-x\left(y+1\right)-4\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(5x+4\left(y+1\right)\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=-\dfrac{5x+4}{4}\end{matrix}\right.\)
Thế vào 1 trong 2 pt ban đầu...