Tìm x,y nguyên x^3 -xy +1 = 2y - x
Tìm x,y nguyên biết \(x^3-y^3=xy+1\)
Rút gọn:
\(A=\dfrac{\sqrt[3]{x^4}+\sqrt[3]{x^2y^2}+\sqrt[3]{y^4}}{\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2}}\)
\(B=\dfrac{\sqrt[3]{xy}\left(\sqrt[3]{y^2}-\sqrt[3]{x^2}\right)+\left(\sqrt[3]{x^4}-\sqrt[3]{y^4}\right)}{\sqrt[3]{x^4}+\sqrt[3]{x^2y^2}-\sqrt[3]{x^3y}}.\sqrt[3]{x^2}\)
\(C=\left(\dfrac{x\sqrt[3]{x}-2x\sqrt[3]{y}+\sqrt[3]{x^2y^2}}{\sqrt[3]{x^2}-\sqrt[3]{xy}}+\dfrac{\sqrt[3]{x^2y}-\sqrt[3]{xy^2}}{\sqrt[3]{x}-\sqrt[3]{y}}\right).\dfrac{1}{\sqrt[3]{x^2}}\)
Phân tích thành nhân tử :
1. \(x^2+xy\left(2y-1\right)=2y^3-2y^2-x\)
2. \(x\sqrt{x^2+y}+y=\sqrt{x^4+x^2}+x\)
3. \(x^4+x^3-11x^2+yx^2+\left(y-12\right)x=12-y\)
4. \(\sqrt{y-1}+2y^2+1=\sqrt{x}+x^2+xy+3y\)
1. \(\left\{{}\begin{matrix}x+xy+y=11\\x^2+y^2-xy-2\left(x+y\right)=-31\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}xy-x+y=-3\\x^2+y^2-x+y+xy=6\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}x^2+4y^2=8\\x+2y=4\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}2+6y=\frac{x}{y}-\sqrt{x-2y}\\\sqrt{x+\sqrt{x-2y}}=x+3y-2\end{matrix}\right.\)
Rút gọn : \(2y\sqrt{x-y}+x\sqrt{\dfrac{1}{x-y}}-x\sqrt{\dfrac{a}{ax-xy}-\sqrt{x^3-x^2y}}\)
(Với x,y >0)
tìm x,y thuộc N* thỏa mãn
x2-xy+y^2=x^2y^2 - 5
tìm x,y thuộc N* thỏa mãn
x2-xy+y^2=x^2y^2 - 5
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
