tap hop cac so nguyen x sao cho 15-Ix-2I=12
tap hop cac so nguyen x thoa man 15-|x-2|=12
Tim tap hop cac so nguyen x sao cho-6<x5 hay tinh tong cac so nguyen trong tap hop do
tap hop cac so nguyen x thoa man 15-/x-2\=12 la
15- |x - 2| = 12
|x - 2| = 3
x - 2 = 3 => x= 5
x - 2= -3 => x = -1
{-1 ; 5}
tap hop cac so nguyen x thoa man -2 ^.Ix+1I=-2^6 la
so phan tu cua tap hop cac so nguyen x thoa man:15-lx-2l bang 12
so nguyen la gi .neu x la so nguyen thi x thuoc tap hop Z kohay tap hop z la tap hop cac so nguyen à
tap hop cac so nguyen to x sao cho 6 chia het cho (x-1)
6 chí hết cho 6,3,2,1
x-1=6 => x=5
x-1=3 => x=2
x-1=2 => x=1
x-1=1 => x=0 không phải số nguyên tố
vậy x cần tìm là 5,2,1
Tim cac so nguyen x sao cho:
a)Ix+2I+Iy+5I=0
b)IIyI+Ix+2II+IxI=0
a) \(\left|x+2\right|+\left|y+5\right|=0\)
\(\Rightarrow\begin{cases}\left|x+2\right|=0\\\left|y+5\right|=0\end{cases}\)
\(\Rightarrow\begin{cases}x+2=0\\y+5=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-2\\y=-5\end{cases}\)
b) \(\left|\left|y\right|+\left|x+2\right|\right|+\left|x\right|=0\)
\(\Rightarrow\begin{cases}\left|\left|y\right|+\left|x+2\right|\right|=0\\\left|x\right|=0\end{cases}\)
\(\Rightarrow\begin{cases}\left|y\right|+\left|x+2\right|=0\\x=0\end{cases}\)
Thay x = 0 vào biểu thức \(\left|y\right|+\left|x+2\right|=0\), ta đc:
\(\left|y\right|+\left|0+2\right|=0\Rightarrow\left|y\right|+2=0\Rightarrow\left|y\right|=-2\Rightarrow y=\phi\)
Vậy \(x=0;y=\phi\)
tap hop cac so nguyen x thoa man -3 x |2x - 3 |= -12