Lời giải:
Sửa đề: Rút gọn \(x+2y-\sqrt{x^2-4xy+4y^2}\) \((x\geq 2y)\)
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Ta có:
\(x+2y-\sqrt{x^2-4xy+4y^2}=x+2y-\sqrt{x^2-2.x.2y+(2y)^2}\)
\(=x+2y-\sqrt{(x-2y)^2}\)
\(=x+2y-|x-2y|=x+2y-(x-2y)=4y\)
(do \(x\geq 2y\Rightarrow |x-2y|=x-2y\) )
\(\left\{{}\begin{matrix}\left(\sqrt{x}-2y\right)\left(1-\dfrac{1}{2y\sqrt{x}}\right)=3\\\left(x+4y^2\right)\left(1+\dfrac{1}{4xy^2}\right)=25\end{matrix}\right.\)
Cho 2 số thực x, y thỏa mãn:
\(\left(x+\sqrt{x^2+2020}\right)\left(2y+\sqrt{4y^2+2020}\right)=2020\)
Tìm GTLN cuẩ biểu thức: B=\(\dfrac{x^2}{2}+4xy+3y^2+x+3y+15\)
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 hpt: \(\left\{{}\begin{matrix}x^2+y^2=2\\\left(x+2y\right)\left(2+3y^2+4xy\right)=27\end{matrix}\right.\)
2/ Cho x,y là các số thực TM: \(1\le y\le2;xy+2\ge2y\). Tìm GTNN:
\(M=\frac{x^2+4}{y^2+1}\)
giải hpt
\(\left\{{}\begin{matrix}x^2+4y^2=5\\\left(x+2y\right)\left(5+4xy\right)=27\end{matrix}\right.\)
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 \(\left\{{}\begin{matrix}\left(x+4y\right)\left(x^2+16y^2\right)=32xy\left(x+4y-3\sqrt{xy}\right)\\\sqrt{3x-1}+6x=\sqrt{8y+3}+8\left(2y+1\right)\end{matrix}\right.\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}2+6y=\frac{x}{y}-\sqrt{x-2y}\\\sqrt{x+\sqrt{x-2y}}=x+3y-2\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^2+y^2+xy+1=4y\\y\left(x+y\right)^2=2x^2-7y+2\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x^2\left(y+1\right)=6y-2\\x^4y^2+2x^2y^2+y\left(x^2+1\right)=12y^2-1\end{matrix}\right.\)
