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
Theo đề, ta có: \(\left\{{}\begin{matrix}2x=3y\\4y=5z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{y}{5}=\dfrac{z}{4}\end{matrix}\right.\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=k\)
=>x=15k; y=10k; z=8k
Ta có: \(3x^2-y^2+z^2=1971\)
\(\Leftrightarrow675k^2-100k^2+64k^2=1971\)
\(\Leftrightarrow k^2=\dfrac{219}{71}\)
Trường hợp 1: \(k=\sqrt{\dfrac{219}{71}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=15\sqrt{\dfrac{219}{71}}\\y=10\sqrt{\dfrac{219}{71}}\\z=8\sqrt{\dfrac{219}{71}}\end{matrix}\right.\)
Trường hợp 2: \(k=-\sqrt{\dfrac{219}{71}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-15\sqrt{\dfrac{219}{71}}\\y=-10\sqrt{\dfrac{219}{71}}\\z=-8\sqrt{\dfrac{219}{71}}\end{matrix}\right.\)