giải hệ pt \(\left\{{}\begin{matrix}y^2+2x^2-3xy-8y+6x-20=0\\4\sqrt{x+2}+\sqrt{16-3y}=x^2+8\end{matrix}\right.\)
Giải phương trình: \(\left\{{}\begin{matrix}y^2+2x^2-3xy-8y+6x-20=0\\4\sqrt{x+2}+\sqrt{16-3y}=x^2+8\end{matrix}\right.\)
Câu hỏi của tiên lê - Toán lớp 10 | Học trực tuyến
Giải các hệ phương trình sau
\(1)\left\{{}\begin{matrix}\sqrt{x+1}=\sqrt{2}\left(8y^2+8y+1\right)\\4\left(x^3-8y^3\right)-6\left(x^2+4y^2\right)+3\left(x+2y\right)-1=0\end{matrix}\right.\)
\(2)\left\{{}\begin{matrix}3\sqrt{17x^2-y^2-6x+4}+x=6\sqrt{2x^2+x+y}-3y+2\\\sqrt{3x^2+xy+1}=\sqrt{x+1}\end{matrix}\right.\)
\(3)\left\{{}\begin{matrix}x^3+\left(2-y\right)x^2+\left(2-3y\right)x=5\left(x+1\right)\\3\sqrt{y+1}=3x^2-14x+14\end{matrix}\right.\)
\(4)\left\{{}\begin{matrix}4x^2=\left(\sqrt{x^2+1}+1\right)\left(x^2-y^3+3y-2\right)\\x^2+\left(y+1\right)^2=2\left(1+\dfrac{1-x^2}{y}\right)\end{matrix}\right.\)
\(5)\left\{{}\begin{matrix}7x^3+y^3+3xy\left(x-y\right)-12x^2+6x-1=0\\y^2+7y-17=9x+2\left(x+6\right)\sqrt{5-2y}\end{matrix}\right.\)
\(6)\left\{{}\begin{matrix}2x^2+3=4\left(x^2-2yx^2\right)\sqrt{3-2y}+\dfrac{4x^2+1}{x}\\\left(2x+1\right)\sqrt{2-\sqrt{3-2y}}=\sqrt[3]{2x^2+x^3}+x+2\end{matrix}\right.\)
Giải các hệ PT sau:
a) \(\left\{{}\begin{matrix}2x^2-3xy=y^2-3x-1\\2y^2-3xy=x^2-3y-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^3-2y=4\\y^3-2x=4\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\sqrt{x+1}-\sqrt{7-y}=4\\\sqrt{y+1}-\sqrt{7-x}=4\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}2x^2=y+\frac{1}{y}\\2y^2=x+\frac{1}{x}\end{matrix}\right.\)
Giải các hệ phương trình sau:a) \(\left\{{}\begin{matrix}\left(2x-y\right)^2-6x+3y=0\\x+2y=0\end{matrix}\right.\);b) \(\left\{{}\begin{matrix}\sqrt{\dfrac{2x-y}{x+y}}+\sqrt{\dfrac{x+y}{2x-y}}=2\\3x+y=14\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
giải hệ phương trình
\(a,\left\{{}\begin{matrix}\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\\y+\frac{y}{\sqrt{x^2-1}}=\frac{35}{12}\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y+1\right)=25\left(y+1\right)\\x^2+xy+2y^2+x-8y=9\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}2x^2+3xy-2y^2-5\left(2x-y\right)=0\\x^2-2xy-3y^2+15=0\end{matrix}\right.\)
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 các hệ phương trình sau:
a, \(\left\{{}\begin{matrix}8x^3y^3+27=18y^3\\4x^2y+6x=y^2\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+4\sqrt{xy}=16\\x+y=10\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x^2-2xy+x-2y+3=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)
Giải hệ pt: (Em xin lỗi vì bài trước em ghi sai đề:()
\(\left\{{}\begin{matrix}x^3+y^3+3xy=1\\\sqrt{\left(4-x\right)\left(13-y\right)}=\dfrac{2x+3y+25}{2x+y+2}\end{matrix}\right.\)
Giải hệ pt
a) \(\left\{{}\begin{matrix}x^3+6x^2y=7\\2y^3+3xy^2=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}6x-xy-2=0\\2\sqrt{\left(x+2\right)\left(3x-y\right)}=y+6\end{matrix}\right.\)