\(\left\{{}\begin{matrix}\left(x+2y\right)^2=5+4xy\\\left(x+2y\right)\left(5+4xy\right)=27\end{matrix}\right.\)
\(\Rightarrow\left(x+2y\right)^3=27\Rightarrow x+2y=3\Rightarrow x=3-2y\)
Thay vào pt đầu:
\(\left(3-2y\right)^2+4y^2-5=0\)
\(\Leftrightarrow8y^2-12y+4=0\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=1\\y=\frac{1}{2}\Rightarrow x=2\end{matrix}\right.\)
giải hpt
a) \(\left\{{}\begin{matrix}x\sqrt{5}-\left(1+\sqrt{3}\right)y=1\\\left(1-\sqrt{3}\right)x+y\sqrt{5}=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}0,2x+0,1y=0,3\\3x+y=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
giải hpt
\(\left\{{}\begin{matrix}x^2+2y^2-3xy-2x+4y=0\\\left(x^2-5\right)^2=2x-2y+5\end{matrix}\right.\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}2y\left(4y^2+3x^2\right)=x^4\left(x^2+3\right)\\2012^x\left(\sqrt{2y-2x+5}-x+1\right)=4024\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\\2x+\frac{1}{x+y}=1\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}x^5-x^4y+x-y=0\left(1\right)\\x^3-3x^2y+4xy^2-4y^3=54\left(2\right)\end{matrix}\right.\)
1. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}x-y=4\\3x+4y=19\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}x-\sqrt{3y}=\sqrt{3}\\\sqrt{3x}+y=7\end{matrix}\right.\)
2. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}2-\left(x-y\right)-3\left(x+y\right)=5\\3\left(x-y\right)+5\left(x+y\right)=-2\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x+y=24\\\dfrac{x}{9}+\dfrac{y}{27}=2\dfrac{8}{9}\end{matrix}\right.\) d, \(\left\{{}\begin{matrix}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2=15}\end{matrix}\right.\)
3. Cho hpt \(\left\{{}\begin{matrix}\left(m+1\right)x-y=3\\mx+y=m\end{matrix}\right.\)
a, Giải hpt khi m=\(\sqrt{2}\)
b, tìm giá trị của m để hpt có nghiệm duy nhất thỏa mãn: x+y>0
Giải hpt
\(\left\{{}\begin{matrix}x^2+4y^2=2\\\left(x-2y\right)\left(1-2xy\right)=4\end{matrix}\right.\)
giải hpt: \(\left\{{}\begin{matrix}\left(x+y\right)^2\left(8x^2+8y^2+4xy-13\right)+5=0\\2x+\frac{1}{x+y}=1\end{matrix}\right.\)
giải HPT
a) \(\left\{{}\begin{matrix}\left(x+3\right)\left(y-5\right)=xy\\\left(2x-y\right)\left(y+15\right)=2xy\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{4x}-3y+4z^2=-2\\\sqrt{3x}+2y-3z^2=1\\-3\sqrt{x}+y+2z^2=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3=30\\x^2y+x\left(1+y+y^2\right)+y=11\end{matrix}\right.\)
giải hpt:
1, \(\left\{{}\begin{matrix}x^3+x^2y+y^3=3\\x^2-xy^2+3y^3=3\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}xy\left(2x+4y\right)=1\\x^3+8y^3+6=32x^2y^2\end{matrix}\right.\)