\(\left\{{}\begin{matrix}17\left(x-y\right)=3xy-2x^2-y^2\\\sqrt{x+3}+\sqrt{10-y}=x^2-7y+11\end{matrix}\right.\)
\(\left\{{}\begin{matrix}17\left(x-y\right)=3xy-2x^2-y^2\\\sqrt{x+3}+\sqrt{10-y}=x^2-7y+11\end{matrix}\right.\)
giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
caau a) binh phuong len ra no x=y tuong tu
c)
ĐK $y \geqslant 0$
Hệ đã cho tương đương với
$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$
Trừ từng vế $2$ phương trình ta được
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
Giải hệ phương trình:
a) \(\left\{{}\begin{matrix}\sqrt{3y^2+13}-\sqrt{15-2x}=\sqrt{x+1}\\y^4-2x^2y+7y^2=\left(x+1\right)\left(8-x\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{x+y}-\sqrt{x-y}=2\\\sqrt{x^2+y^2+1}-\sqrt{x^2-y^2}=3\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\sqrt{2x+y+1}-\sqrt{x+y}=3\\\sqrt{3\left(x+y\right)^2+1}+\sqrt{x-5}=5\end{matrix}\right.\)
1.\(\left\{{}\begin{matrix}x\left(x-2\right)\left(2x-y\right)=6\\\left(x-3\right)^2+2y=10\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{y}}=2\\\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{x}}=2\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=8\\x\left(x+1\right)+y\left(y+1\right)+xy=17\end{matrix}\right.\)
\(\Rightarrow\left|a\right|\le1\),\(\left|b\right|\le1\),\(\left|c\right|\le1\)
\(\Rightarrow1-a\ge0\)tương tự 1-b,1-c............
\(\Rightarrow\left(1\right)\ge0\)
dấu = khi a=1b=0c=0 và hoán vị
Đang nổi cơn làm biếng mà nhìn thấy hệ còn buồn ngủ hơn:
a/ \(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-2x\right)\left(2x-y\right)=6\\x^2-2x-2\left(2x-y\right)=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-2x=a\\2x-y=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}ab=6\\a-2b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=6\\a=2b+1\end{matrix}\right.\)
\(\Rightarrow b\left(2b+1\right)=6\Leftrightarrow2b^2+b-6=0\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: ...
\(\Leftrightarrow\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{y}}-\sqrt{2-\frac{1}{x}}=0\)
\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\frac{1}{x}-\frac{1}{y}}{\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}}=0\)
\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{y-x}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)
\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)
\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\frac{1}{\sqrt{xy}}+\frac{\sqrt{y}+\sqrt{x}}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}\right)=0\)
\(\Leftrightarrow\sqrt{y}=\sqrt{x}\Rightarrow x=y\)
Thay vào pt đầu:
\(\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}=2\)
\(\Leftrightarrow\frac{1}{x}+2-\frac{1}{x}+2\sqrt{\frac{2}{x}-\frac{1}{x^2}}=4\)
\(\Leftrightarrow\sqrt{\frac{2}{x}-\frac{1}{x^2}}=1\)
\(\Leftrightarrow\frac{2}{x}-\frac{1}{x^2}=1\)
\(\Leftrightarrow\left(\frac{1}{x}-1\right)^2=0\)
c/ \(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2+x+y+xy=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\\left(x+y\right)^2-2xy=10\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\) ta được:
\(\left\{{}\begin{matrix}a+b=7\\a^2-2b=10\end{matrix}\right.\) \(\Rightarrow a^2+2a-24=0\Rightarrow\left[{}\begin{matrix}a=4;b=3\\a=-6;b=13\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)
1) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}2x^2+y^2-3xy+3x-2y+1=0\\4x^2-y^2+x+4=\sqrt{2x+y}+\sqrt{x+4y}\end{matrix}\right.\)
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 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 hệ phương trình:
a, \(\left\{{}\begin{matrix}\sqrt{2x+3}+\sqrt{4-y}=4\\\sqrt{2y+3}+\sqrt{4-y}=4\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}2y=\dfrac{y^2+1}{x^2}\\2x=\dfrac{x^2+1}{y^2}\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}2x^2+y^2-3xy=12\\2\left(x+y\right)^2-y^2=14\end{matrix}\right.\)
Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!