Với giá trị nào của a thì \(\left\{{}\begin{matrix}x+y=a+2\\3x+5y=2a\end{matrix}\right.\) có nghiệm là các số nguyên
a)\(\left\{{}\begin{matrix}8y-x=4\\2x-21y=2\end{matrix}\right.\) b)\(\left\{{}\begin{matrix}x+y=-2.\left(x-1\right)\\7x+3y=x+y+5\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x^2-y^2=3\left(x-y\right)\\xy=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\sqrt{y}+y\sqrt{x}=6\\x^2y+y^2x=20\end{matrix}\right.\)
giải các hpt sau: a)\(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}y=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{3x}{4}+\dfrac{2y}{5}=2,3\\x-\dfrac{3y}{5}=0,8\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)cíu zới
Ghpt:
a) \(\left\{{}\begin{matrix}x^2+2y^2=2x-2xy+1\\3x^2+2xy-y^2=2x-y+5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4xy+4x^2+4y^2+\dfrac{3}{\left(x+y\right)^2}=7\\2x+\dfrac{1}{x+y}=3\end{matrix}\right.\)
giải các phương trình sau a)\(\left\{{}\begin{matrix}3\left(x-1\right)-\sqrt{1-2y}=1\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{x^2-2x+1}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}2\left(x^2-x\right)+\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
giải hệ phương trình (theo 4 cách):
a/ \(\left\{{}\begin{matrix}\sqrt{5}x-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
b/ \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}ax+2ay=a+1\\x+\left(a+1\right)y=2\end{matrix}\right.\)
tìm a để hệ có nghiệm duy nhất (x;y)
Hai số a,b thỏa mãn \(\left\{{}\begin{matrix}a,b>0\\\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)\ge4\end{matrix}\right.\)
Chứng minh \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\ge2\)