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...
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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, 2017-|x-2017| = x
=> |x - 2017| = 2017 - x
Th1: x \(\ge\)2017
=> x - 2017 = 2017 - x
=> x + x = 2017 + 2017
=> x = 2017 (thỏa mãn)
Th2: x < 2017
=> x - 2017 = -2017 + x
=> x - x = -2017 + 2017
=> 0 = 0
Vậy x = 2017
b, Vì \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\\\left(3y-7\right)^{2020}\ge0\\\left|x+y+z\right|\ge0\end{cases}\forall x,y,z}\)
\(\Rightarrow\left(2x-5\right)^{2018}+\left(3y-7\right)^{2020}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x-5\right)^{2018}+\left(3y-7\right)^{2020}+\left|x+y+z\right|=0\)
Do đó \(\hept{\begin{cases}\left(2x-5\right)^{2018}=0\\\left(3y-7\right)^{2020}=0\\\left|x+y+z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y-7=0\\x+y+z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{7}{3}\\z=\frac{-29}{6}\end{cases}}}\)
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