Giai hệ phương trình:
\(\left\{{}\begin{matrix}x^3+2y^2-4y+3=0\\x^2+x^2y^2-xy=0\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}x^3+xy^2+3\left(x-2y\right)=0\\x^2+xy=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3+xy^2+3\left(x-2y\right)=0\\x^2+xy=3\end{matrix}\right.\)\(\Rightarrow x^3+xy^2+\left(x^2+xy\right)\left(x-2y\right)=0\)\(\Leftrightarrow x^3+xy^2+x^3-x^2y-2xy^2=0\Leftrightarrow2x^3-x^2y-xy^2=0\)\(\Leftrightarrow x\left(2x+y\right)\left(x-y\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=-2x\\x=y\end{matrix}\right.\)
+) \(x=0\Rightarrow0y=3\)(vô nghiệm)
+) y=-2x \(\Rightarrow x^2-2x^2=3\Leftrightarrow-x^2=3\)(vô nghiệm)
+) x=y\(\Rightarrow2x^2=3\Leftrightarrow x^2=\dfrac{3}{2}\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt{\dfrac{3}{2}}\\x=y=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)
giải hệ phương trình \(\left\{{}\begin{matrix}x^3-12y^3+xy\left(16y-7x\right)=0\\\sqrt{x-2y}+\sqrt{x+2y}=2\end{matrix}\right.\)
\(x^3-7x^2y+16xy^2-12y^3=0\)
\(\Leftrightarrow\left(x-3y\right)\left(x-2y\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=3y\end{matrix}\right.\)
Thế xuống pt dưới giải đơn giản
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 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 hệ phương trình: \(\left\{{}\begin{matrix}x^3-y^3+2x^2+4y^2+5-0\\x^2+2y^2+4x-13y+7=0\end{matrix}\right.\)
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:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
Giải hệ bằng phương pháp phân tích nhân tử
a) \(\left\{{}\begin{matrix}x^2+2y=xy+4\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-2xy+x+y=0\\x^4-4x^2y+3x^2+y^2=0\end{matrix}\right.\)