tim x dua vao quan he uoc boi:tim so tu nhien x sao cho x-1 la uoc cua 12tim so tu nhien x sao cho 2x+1 la uoc cua 28tim so tu nhien x sao cho x+15 la boi cua x+3tim cac so nguyen x,y sao cho (x+1)(y-2)=3tim so nguyen x sao cho(x+2).(y-1)=2tim so nguyen to x vua la uoc cua 275 vua la uoc cua 180tim so nguyen to x,y biet x+y=12 va UCLL (x:y)=5tim so tu nhien x,y biet x+y=32 va UCLL (x:y)=8tim so tu nhien x biet x chia het cho10; xchia het cho12; x chia het cho15 va 100<x<150tim so x nho nhat khac 0b...
Đọc tiếp
tim x dua vao quan he uoc boi:
tim so tu nhien x sao cho x-1 la uoc cua 12
tim so tu nhien x sao cho 2x+1 la uoc cua 28
tim so tu nhien x sao cho x+15 la boi cua x+3
tim cac so nguyen x,y sao cho (x+1)(y-2)=3
tim so nguyen x sao cho(x+2).(y-1)=2
tim so nguyen to x vua la uoc cua 275 vua la uoc cua 180
tim so nguyen to x,y biet x+y=12 va UCLL (x:y)=5
tim so tu nhien x,y biet x+y=32 va UCLL (x:y)=8
tim so tu nhien x biet x chia het cho10; xchia het cho12; x chia het cho15 va 100<x<150
tim so x nho nhat khac 0b biet x chia het cho 24 va 30
40 chia het cho x . 56 chia het cho x va x>6
a/ Với mọi x,y ta có :
\(\hept{\begin{cases}\left|x\right|\ge0\\\left|y\right|\ge0\end{cases}}\)
\(\Leftrightarrow\left|x\right|+\left|y\right|\ge0\)
Mặt khác : \(\left|x\right|+\left|y\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}\)
Vậy ...
b/ Với mọi x,y ta có :
\(\hept{\begin{cases}\left|x-1\right|\ge0\\\left|y\right|\ge0\end{cases}}\)
\(\Leftrightarrow\left|x-1\right|+\left|y\right|\ge0\)
Mà \(\left|x-1\right|+\left|y\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x-1\right|=0\\\left|y\right|=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=1\\y=0\end{cases}}\)
Vậy ...
b/ Với mọi x,y ta có :
\(\hept{\begin{cases}\left|x+2\right|\ge0\\\left|y-1\right|\ge0\end{cases}}\)
\(\Leftrightarrow\left|x+2\right|+\left|y-1\right|\ge0\)
Mà \(\left|x+2\right|+\left|y-1\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+2\right|=0\\\left|y-1\right|=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)
Vậy ..
a) |x|+|y|=0
\(\left|x\right|\ge0;\left|y\right|\ge0\Rightarrow\left|x\right|+\left|y\right|=0\)
\(\Leftrightarrow x=0;y=0\)
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