ghpt \(\left\{{}\begin{matrix}x^2+2x=\dfrac{4y}{x}+\dfrac{3}{y^2}\\x^2-\dfrac{2y^2}{x^2}=\dfrac{4y}{x}-6y\end{matrix}\right.\)
Ghpt:
a) \(\left\{{}\begin{matrix}x^2+2y^2=2x-2xy+1\\3x^2+2xy-y^2=2x-y+5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4xy+4x^2+4y^2+\dfrac{3}{\left(x+y\right)^2}=7\\2x+\dfrac{1}{x+y}=3\end{matrix}\right.\)
Giải hệ phương trình
a)\(\left\{{}\begin{matrix}x+y=\dfrac{x-3}{2}\\x+2y=\dfrac{2-4y}{15}\end{matrix}\right.\) b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\) d)\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{9}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x^2+3xy-2y^2=2x+4y\\\dfrac{x^2\left(2x-y\right)}{x+2y}=\dfrac{1}{2}\end{matrix}\right.\)
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 phương trình:
a) \(\left\{{}\begin{matrix}4x^3+y^2-2y+5=0\\x^2+x^2y^2-4y+3=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{2x^2}{x^2+1}=y\\\dfrac{3y^3}{y^4+y^2+1}=z\\\dfrac{4z^4}{z^6+z^4+z^2+1}=x\end{matrix}\right.\)
Pt đầu chắc là sai đề (chắc chắn), bạn kiểm tra lại
Với pt sau:
Nhận thấy một ẩn bằng 0 thì 2 ẩn còn lại cũng bằng 0, do đó \(\left(x;y;z\right)=\left(0;0;0\right)\) là 1 nghiệm
Với \(x;y;z\ne0\)
Từ pt đầu ta suy ra \(y>0\) , từ đó suy ra \(z>0\) từ pt 2 và hiển nhiên \(x>0\) từ pt 3
Do đó:
\(\left\{{}\begin{matrix}y=\dfrac{2x^2}{x^2+1}\le\dfrac{2x^2}{2x}=x\\z=\dfrac{3y^3}{y^4+y^2+1}\le\dfrac{3y^3}{3\sqrt[3]{y^4.y^2.1}}=y\\x=\dfrac{4z^4}{z^6+z^4+z^2+1}\le\dfrac{4z^4}{4\sqrt[4]{z^6z^4z^2}}=z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y\le x\\z\le y\\x\le z\end{matrix}\right.\) \(\Rightarrow x=y=z\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)
Vậy nghiệm của hệ là \(\left(x;y;z\right)=\left(0;0;0\right);\left(1;1;1\right)\)
\(Ghpt:\left\{{}\begin{matrix}x-\dfrac{1}{x^3}=y-\dfrac{1}{y^3}\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x^3}=y-\dfrac{1}{y^3}\left(1\right)\\\left(x-4y\right)\left(2x-y+4\right)=-36\left(2\right)\end{matrix}\right.\)
\(Đk:\left\{{}\begin{matrix}x,y\ne0\\x\ne4y\\2x\ne y-4\end{matrix}\right.\)
\(x-\dfrac{1}{x^3}=y-\dfrac{1}{y^3}\)
\(\Rightarrow x-y+\dfrac{1}{y^3}-\dfrac{1}{x^3}=0\)
\(\Rightarrow x-y+\dfrac{x^3-y^3}{x^3y^3}=0\)
\(\Rightarrow x-y+\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^3y^3}=0\)
\(\Rightarrow\left(x-y\right).\dfrac{x^2+xy+y^2+x^3y^3}{x^3y^3}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x^2+xy+y^2+x^3y^3=0\end{matrix}\right.\)
Với \(x=y\) . Thay vào (2) ta được:
\(\left(x-4x\right)\left(2x-x+4\right)=-36\)
\(\Leftrightarrow-3x.\left(x+4\right)=-36\)
\(\Leftrightarrow x\left(x+4\right)=12\)
\(\Leftrightarrow x^2+4x-12=0\)
\(\Leftrightarrow\left(x+2\right)^2-16=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\Rightarrow y=2\\x=-6\Rightarrow y=-6\end{matrix}\right.\)
Với \(x^2+xy+y^2+x^3y^3=0\) . Ta sẽ chứng minh trường hợp này vô nghiệm.
Có: \(\left(x+y\right)^2+x^3y^3-xy=0\)
\(\Rightarrow\left(x+y\right)^2+xy\left(xy+1\right)\left(xy-1\right)=0\left(3\right)\)
Với \(xy>1\Rightarrow VT\left(3\right)>0\Rightarrow ptvn\)
Với \(xy=1\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
\(\Rightarrow x^2=-1\Rightarrow ptvn\)
Với \(1>xy\ge0\Rightarrow xy\left(xy+1\right)\left(xy-1\right)\le0\) (có thể xảy ra).
Với \(0>xy>-1\Rightarrow VT\left(3\right)>0\Rightarrow ptvn\)
Với \(xy< -1\Rightarrow xy\left(xy-1\right)\left(xy+1\right)\le0\) (có thể xảy ra).
Vì \(x,y\ne0\) nên ta có: \(\left[{}\begin{matrix}1>xy>0\\xy< -1\end{matrix}\right.\left('\right)\)
\(\left(2\right)\Rightarrow2x^2-xy+4x-8xy+4y^2-16y=-36\)
\(\Rightarrow2x^2+4x+4y^2-16y+36=9xy\)
\(\Rightarrow2\left(x^2+2x+1\right)+4\left(y^2-4y+4\right)+18=9xy\)
\(\Rightarrow2\left(x+1\right)^2+4\left(y-2\right)^2+18=9xy>18\)
\(\Rightarrow xy>2\left(''\right)\)
Từ \(\left('\right),\left(''\right)\) suy ra hệ vô nghiệm.
Vậy hệ phương trình đã cho có nghiệm \(\left(x,y\right)\in\left\{\left(2;2\right),\left(-6;-6\right)\right\}\)
GHPT: \(\left\{{}\begin{matrix}x^2+1+y^2+xy=4y\\x+y-2=\dfrac{y}{x^2+1}\end{matrix}\right.\)
- Với \(y=0\) không phải nghiệm
- Với \(y\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2+1}{y}+x+y=4\\x+y-2=\dfrac{y}{x^2+1}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x^2+1}{y}+2=4-\dfrac{y}{x^2+1}\)
Đặt \(\dfrac{x^2+1}{y}=t\Rightarrow t=2-\dfrac{1}{t}\Leftrightarrow t^2-2t+1=0\)
\(\Rightarrow t=1\Rightarrow\dfrac{x^2+1}{y}=1\Rightarrow\dfrac{y}{x^2+1}=1\)
Thế xuống pt dưới: \(x+y-2=1\Rightarrow x=3-y\)
Thế vào pt trên: \(\left(3-y\right)^2+1+y^2+y\left(3-y\right)=4y\)
\(\Leftrightarrow...\)
1, \(\left\{{}\begin{matrix}x^3+2y^2-4y+29=0\\x^2+x^2y^2-18y=0\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+2y^2-4y+10=0\\x^2+x^2y^2-16y+12=0\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x,y>0\\x+y=7\\\dfrac{9}{x}+\dfrac{16}{y}=7\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}x,y>0\\x+y=4\\\dfrac{4}{x}+\dfrac{9}{y}\le4\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}x^3+y^2=\dfrac{211}{27}\\x^2+y^2+xy-3x-4y+4=0\end{matrix}\right.\)
6, \(\left\{{}\begin{matrix}x^4+81y^2=697\\x^2+9y^2+3xy-9x-36y+36=0\end{matrix}\right.\)
Giải hệ phương trình
a)\(\left\{{}\begin{matrix}\dfrac{2x-1}{x+2}-\dfrac{5}{y-1}=-\dfrac{14}{3}\\\dfrac{3}{x+2}+\dfrac{\left(2y+3\right)}{y-1}=8\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{2x}{1-x}+\dfrac{3}{y+2}=-\dfrac{2}{5}\\\dfrac{5}{1-x}-\dfrac{4y}{y+2}=\dfrac{1}{10}\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x+4-5}{x+2}-\dfrac{5}{y-1}=-\dfrac{14}{3}\\\dfrac{3}{x+2}+\dfrac{2y-2+5}{y-1}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{x+2}-\dfrac{5}{y-1}=-\dfrac{14}{3}-2=-\dfrac{20}{3}\\\dfrac{3}{x+2}+\dfrac{5}{y-1}=6\end{matrix}\right.\)
=>x+2=3 và y-1=1
=>x=1 và y=2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2x}{x-1}+\dfrac{3}{y+2}=\dfrac{-2}{5}\\\dfrac{-5}{x-1}-\dfrac{4y}{y+2}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2x+2-2}{x-1}+\dfrac{3}{y+2}=\dfrac{-2}{5}\\\dfrac{-5}{x-1}-\dfrac{4y+8-8}{y+2}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{x-1}+\dfrac{3}{y+2}=-\dfrac{2}{5}+2=\dfrac{8}{5}\\\dfrac{-5}{x-1}+\dfrac{8}{y+2}=\dfrac{1}{10}-4=-\dfrac{39}{10}\end{matrix}\right.\)
=>x-1=-2/49 và y+2=-5/79
=>x=47/49 và y=-5/79-2=-163/79