1. (-34)+24+(-7)+27=-10+20=10

2.99+(-100)+101=(99+101)+(-100)=200-100=100

3.(-75)+69+(-25)+131=(69+131)-(75+25)=200-100=100

ta có:\(5x-3y=4y\Rightarrow5x=7y\Leftrightarrow x=\dfrac{7y}{5}\)(1)

mà \(4y=3z+10x\Rightarrow4y=3z+14y\)

\(\Leftrightarrow-10y=3z\Leftrightarrow z=\dfrac{-10y}{3}\) (2)

thay (1), (2) vào 3x+2y+z=989, ta co:

\(\dfrac{21y}{5}+2y-\dfrac{10y}{3}=989\Leftrightarrow\dfrac{43y}{15}=989\)

\(\Leftrightarrow y=345\)

thay y=345 vào (1), (2) ta dc: \(\left\{{}\begin{matrix}x=\dfrac{7\times345}{5}=483\\z=\dfrac{-10\times345}{3}=-1150\end{matrix}\right.\)

vậy \(\left\{{}\begin{matrix}x=483\\y=345\\z=-1150\end{matrix}\right.\)

1.\(A=1+2+...+13+14\)

\(A=\left(1+14\right)+\left(2+13\right)+...+\left(7+8\right)\)

\(A=15\times7=105\)

vậy A chia hết cho các ước của 105

ta có:\(\dfrac{x+y+1}{z}+1=\dfrac{x+z+2}{y}+1=\dfrac{y+z-3}{z}+1=\dfrac{1}{x+y+z}+1\Rightarrow\dfrac{x+y+z+1}{z}=\dfrac{x+y+z+2}{y}=\dfrac{x+y+z-3}{z}=\dfrac{1}{x+y+z}+1\)

\(\Rightarrow\dfrac{3\times\left(x+y+z\right)}{x+y+z}=\dfrac{1}{x+y+z}+1\Rightarrow3=\dfrac{1}{x+y+z}+1\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)mà: \(\left\{{}\begin{matrix}\dfrac{x+y+z+1}{x}=3\\\dfrac{x+y+z+2}{y}=3\\\dfrac{x+y+z-3}{z}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)

vậy........

ta có: \(7\times9\times11-2\times3\times7=3\times7\times\left(3\times11-2\right)=3\times7\times31\)

vậy hiệu này là hợp số

gọi số tự nhiên cần tìm là a

ta có:\(\left\{{}\begin{matrix}a⋮8\\a⋮10\\a⋮15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a⋮8\\a⋮5\\a⋮3\end{matrix}\right.\)

mà Ư(8;5;3)=1 \(\Rightarrow a=k.8.3.5=120k\)

mặt khác: 100<a<2000 => \(a\in\left\{120;240;360;480;600;720;840;960;1080;1200;1320;1440;1560;1680;1800;1920\right\}\)

vì(x+3)(2y-5)=34 \(\Rightarrow x+3=\dfrac{34}{2y-5}\)

mà \(x,y\in N\Rightarrow34⋮2y-5\Rightarrow2y-5\in\left\{1;2;17;34\right\}\Rightarrow y\in\left\{3;\dfrac{7}{2};11;\dfrac{39}{2}\right\}\)

\(\Rightarrow\left[{}\begin{matrix}y=3\\y=11\end{matrix}\right.\)(vì \(y\in N\))

với y=3 =>x=31(TM)

với y=11 =>x=-1(loại)

\(M=\left\{2;5\right\};N=\left\{2;5\right\}\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x-5< 6\\x-5>-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 11\\x>-1\end{matrix}\right.\Leftrightarrow-1< x< 11\)

1. vì (2x-1)(y-1)=29 mà \(x,y\in N\)\(\Rightarrow\left\{{}\begin{matrix}2x-1>0\\y-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\y>1\end{matrix}\right.\)

ta có:\(\left(2x-1\right)\left(y-1\right)=29\Rightarrow2x-1=\dfrac{29}{y-1}\)

vì: \(x\in N\Rightarrow\dfrac{29}{y-1}\in N\)

\(\Rightarrow29⋮y-1\Rightarrow y\in\left\{2;30\right\}\)

với y=2 => x=15

với y=30 => x=1