Giải hệ phương trình
a,\(\left\{{}\begin{matrix}1+4x^2y^2+5xy=10x^2\\1-2xy+3y=2x\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\sqrt{x+1}\sqrt{y+1}=\sqrt{4-x+5y}\\x^2+y+2=\sqrt{5\left(2x-y+1\right)}+\sqrt{3x+2}\end{matrix}\right.\)
a. ĐKXĐ: ..
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}-\sqrt{2\left(x+y\right)}=4\\x+2y+\dfrac{2\sqrt{\left(x+y\right)\left(2x+5y\right)}}{3}=24\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}=a\ge0\\\sqrt{2\left(x+y\right)}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=4\\\dfrac{a^2+b^2}{6}+\dfrac{ab}{3}=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left(a+b\right)^2=144\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left[{}\begin{matrix}a+b=12\\a+b=-12\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(8;4\right)\\\left(a;b\right)=\left(-4;-8\right)\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\left(2x+5y\right)=64\\2\left(x+y\right)=16\end{matrix}\right.\) \(\Leftrightarrow...\)
b.
Thế pt trên xuống dưới:
\(x^4+6y^4=\left(x+2y\right)\left(x^3+3y^3-2xy^2\right)\)
\(\Leftrightarrow2x^3y-2x^2y^2-xy^3=0\)
\(\Leftrightarrow xy\left(2x^2-2xy-y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\y=-\left(1+\sqrt{3}\right)x\\y=\left(-1+\sqrt{3}\right)x\end{matrix}\right.\)
Thế vào pt đầu ...
Đề cho hơi xấu, nếu pt đầu là \(x^3+3y^3-2x^2y=1\) thì đẹp hơn nhiều
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:
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:
1, \(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^2\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}=-x^2\left(x^4+1-2x^2-2xy^2\right)\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{x}\left(3\sqrt{x}-y\right)+x\sqrt{x}=3y+\sqrt{y-1}\\3xy^2+4=4x^2+2y+x\end{matrix}\right.\)
Giải các hệ phương trình sau bằng phương pháp cộng đại số
a) \(\left\{{}\begin{matrix}x-y=1\\3x+2y=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}3x+5y=10\\2x+3y=3\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\sqrt{5x}+y=2\\\left(1-\sqrt{5}\right)x-y=-1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\sqrt{3x}-y=1\\3x+\sqrt{3y}=3\end{matrix}\right.\)
Giải hệ phương trình sau bằng phương pháp thế
a)
\(\left\{{}\begin{matrix}\sqrt{5}+2)x+y=3-\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\)
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) \(\left\{{}\begin{matrix}\sqrt{2x+y+1}-\sqrt{x+y}=3\\\sqrt{3\left(x+y\right)^2+1}+\sqrt{x-5}=5\end{matrix}\right.\)
Giải hệ
a) \(\left\{{}\begin{matrix}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(6x+5\right)\sqrt{2x+1}-2y-3y^3=0\\y+\sqrt{x}=\sqrt{2x^2+4x-23}\end{matrix}\right.\)
Giải bất pt
\(\dfrac{9}{\left|x-5\right|-3}\ge\left|x-2\right|\)