Giải hpt sau :
\(\sqrt{ }\)x2+y2 + \(\sqrt{ }\)2xy = 2\(\sqrt{ }\)2
\(\sqrt{ }\)x + \(\sqrt{ }\)y = 4
Giải hpt :
\(\left\{{}\begin{matrix}x+2y-1-2\sqrt{2xy+x-4y-2}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2\\y\ge-\dfrac{1}{2}\end{matrix}\right.\)
Ta có \(\left\{{}\begin{matrix}x+2y-1-2\sqrt{2xy+x-4y-2}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)+\left(2y+1\right)-2\sqrt{\left(x-2\right)\left(2y+1\right)}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2}-\sqrt{2y+1}\right)^2=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\4\sqrt{2y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\2y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)
giải hpt
\(\left\{{}\begin{matrix}2\sqrt{2xy-y}+2x+y=10\\\sqrt{3y+4}-\sqrt{2y+1}+2\sqrt{2x-1}=3\end{matrix}\right.\)
Trừ hai vế của pt đầu cho 1 ta được:
\(\left(\sqrt{y}+\sqrt{2x-1}\right)^2=9\Leftrightarrow\sqrt{y}+\sqrt{2x-1}=3\)
Thay vào (2) ta được:\(\sqrt{3y+4}-\sqrt{2y+1}+3-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{3y+4}-2\sqrt{y}\right)+\left(3-\sqrt{2y+1}\right)=0\)
\(\Leftrightarrow\left(4-y\right)\left(\frac{1}{\sqrt{3y+4}+2\sqrt{y}}+\frac{2}{\sqrt{2y+1}+3}\right)=0\)
\(\Leftrightarrow y=4\Rightarrow x=1\)
Vậy hệ có nghiệm duy nhất (x;y)=(1;4)
Giải hpt :
\(\hept{\frac{\sqrt{\frac{x^2+y^2}{2}}+\sqrt{\frac{x^2+xy+y^2}{3}}=x+y}{x\sqrt{2xy+5x+3}=4xy-5x-3}}\)
Giải hpt: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=2\\\sqrt{x+3}+\sqrt{y+3}=4\end{matrix}\right.\)
Giải hpt \(\left\{{}\begin{matrix}\sqrt{y+3x}+\sqrt{2x+7y}=\sqrt{5x-y}+3\sqrt{x}\\x-4-\sqrt{y-2}=\sqrt{x^3-10x^2+33x-34}-\sqrt{y^3-9y^2+24y-16}\end{matrix}\right.\)
\(ĐK:x\ge0;y\ge2;5x-y\ge0\\ PT\left(1\right)\Leftrightarrow\sqrt{y+3x}-\sqrt{5x-y}+\sqrt{2x+7y}-3\sqrt{x}=0\\ \Leftrightarrow\dfrac{2y-2x}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7y-7x}{\sqrt{2x+7y}+3\sqrt{x}}=0\\ \Leftrightarrow\left(y-x\right)\left(\dfrac{2}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7}{\sqrt{2x+7y}+3\sqrt{x}}\right)=0\\ \Leftrightarrow x=y\left(\dfrac{2}{\sqrt{y+3x}+\sqrt{5x-y}}+\dfrac{7}{\sqrt{2x+7y}+3\sqrt{x}}>0\right)\)
Thay vào \(PT\left(2\right)\Leftrightarrow x-4+\sqrt{x-2}=\sqrt{x^3-10x^2+33x-34}-\sqrt{x^3-9x^2+24x-16}\)
\(\Leftrightarrow\dfrac{x^2-9x+18}{x-4+\sqrt{x-2}}=\dfrac{-x^2+9x-18}{\sqrt{x^3-10x^2+33x-34}+\sqrt{x^3-9x^2+24x-16}}\\ \Leftrightarrow\left(x^2-9x+18\right)\left(\dfrac{1}{x-4+\sqrt{x-2}}+\dfrac{1}{\sqrt{x^3-10x^2+33x-34}+\sqrt{x^3-9x^2+24x-16}}\right)=0\\ \Leftrightarrow x^2-9x+18=0\left(\text{ngoặc lớn luôn }>0,\forall x\ge2\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=6\end{matrix}\right.\)
Vậy ...
giải hpt:
\(\hept{\begin{cases}x+\frac{2xy}{\sqrt[3]{x^2-2x+9}}=x^2+y\\y+\frac{2xy}{\sqrt[3]{y^2-2y+9}}=y^2+x\end{cases}}\)
EZ game
Xét x=y=0
Xét x và y khác 0
Cộng từng vế hai phương trình
Đánh giá VP >= VT
1, Giải hpt : \(\sqrt{x}+\sqrt{6-y}=2\sqrt{3}\)
\(\sqrt{y}+\sqrt{6-x}=2\sqrt{3}\)
Giải hpt sau:
\(\left(x-1\right)\sqrt{y}+\left(y-1\right)\sqrt{x}=\sqrt{2xy}\)
\(x\sqrt{y-1}+y\sqrt{x-1}=xy\)
giải hpt: \(\hept{\begin{cases}\sqrt{x+y}-\sqrt{x-y}=2\\\sqrt{x^2+y^2}+\sqrt{x^2-y^2}=4\end{cases}}\)