Tìm x,y , biết :
a,\(\hept{\begin{cases}x\left(x+y\right)=\frac{1}{48}\\y\left(x+y\right)=\frac{1}{24}\end{cases}}\)
b,\(\hept{\begin{cases}x\left(x-y\right)=\frac{3}{10}\\y\left(x+y\right)=-\frac{3}{10}\end{cases}}\)
Tìm 2 số nguyên dương x, y biết:
\(\hept{\begin{cases}\left(x+1\right)⋮y\\\left(y+1\right)⋮x\\x>y>1\end{cases}}\)
Tìm x biết
\(\hept{\begin{cases}x\left(x+y\right)=\frac{-1}{12}\\y\left(x+y\right)=\frac{1}{9}\end{cases}}\)
Tìm \(x,y,z\in Z\) sao cho:
\(\hept{\begin{cases}y\left(x+y+z\right)=18\\x\left(x+y+z\right)=-12\\z\left(x+y+z\right)=3\end{cases}}\)
tìm x,y,z thuộc Z sao cho
\(\hept{\begin{cases}x+y=\left(-8\right)\\y+z=4\\z-x=\left(-6\right)\end{cases}}\)
Giải hệ phương trình (ẩn số x,y,z):\(\hept{\begin{cases}x+y+z=6\left(1\right)\\x^2+y^2+z^2=18\left(2\right)\\\sqrt{x}+\sqrt{y}+\sqrt{z}=4\left(3\right)\end{cases}.}\)
bai 1: Tim x biet
\(\hept{\begin{cases}x-y=\frac{3}{10}\\y\left(x-y\right)=-\frac{3}{50}\end{cases}}\)
bai 2: Tim x, y biet:
x+\(\left(-\frac{31}{12}\right)^2\)=\(\left(\frac{49}{12}\right)^2\)-x=y2
Bai 9: Tim x,y,z biet:
(x-1)2+(x+y)2+(xy-z)2=0
\(\left(2x+3\right)^2+\left(3x-2\right)^4=0\) vì \(\left(2x+3\right)^2\ge0;\left(3x-2\right)^4\ge0\) nên\(\Rightarrow\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x+3=0\\3x-2=0\end{cases}}}\) \(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}\)
Cho x;y thỏa mãn \(\hept{\begin{cases}\left|6x-3y\right|\le9\\\left|2x-6y\right|\le2\end{cases}}\)
Tính Max P=\(x^4+xy+y^4\)
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