a) Ta có: \(\left|x+4\right|< 3\)
\(\Rightarrow\left|x+4\right|\in\left\{0;1;2\right\}\)
\(\Rightarrow x+4\in\left\{0;\pm1;\pm2\right\}\)
Ta có bảng
x+4 | 0 | 1 | -1 | 2 | -2 |
x | -4 | -3 | -5 | -2 | -6 |
Vậy...
b) ta có: \(\left|x-14+17\right|+\left|y+10-12\right|\le0\)
Mà \(\left|x-14+17\right|+\left|y+10-12\right|\ge0\)
\(\Rightarrow\left|x-14+17\right|+\left|y+10-12\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x-14+17\right|=0\\\left|y+10-12\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x-14+17=0\\y+10-12=0\end{cases}\Rightarrow}\hept{\begin{cases}x=14-17\\y=-10+12\end{cases}\Rightarrow}\hept{\begin{cases}x=-3\\y=2\end{cases}}}\)
Vậy ....
hok tốt!!
á) | x + 4 | < 3
Ta lại có | x + 4 | ≥ 0 \(\forall\) x ∈ Z
Mà x ∈ Z
<=> | x + 4 | ∈ { 0 ; 1 ; 2 }
\(\Leftrightarrow x+4\in\left\{0;1;-1;2;-2\right\}\)
<=> x ∈ { - 4 ; - 3 ; - 7 ; - 2 ; - 6 }
Vậy ...
b) | x - 14 + 17 | + | y + 10 - 12 | ≤ 0
<=> | x + 3 | + | y - 2 | ≤ 0
+) Lại có \(\hept{\begin{cases}\left|x+3\right|\text{≥}0\\\left|y-2\right|\text{≥}0\end{cases}\forall x;y}\)
<=> | x + 3 | + | y - 2 | ≥ 0 \(\forall\) x ; y
Do đó để | x + 3 | + | y - 2 | ≤ 0 thì \(\hept{\begin{cases}\left|x+3\right|=0\\\left|y-2\right|=0\end{cases}}\)
<=> \(\hept{\begin{cases}x+3=0\\y-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-3\\y=2\end{cases}}\)
Vậy ..... <=> x = - 3 và y = 2
Bài giải
a, \(\left|x+4\right|< 3\)
\(\Rightarrow\text{ }-3< x+4< 3\)
\(\Rightarrow\text{ }-7< x< -1\)
\(\Rightarrow\text{ }x\in\left\{-6\text{ ; }-5\text{ ; }-4\text{ ; }-3\text{ ; }-2\right\}\)
b, \(\left|x-14+17\right|+\left|y+10-12\right|\le0\)
\(\left|x+3\right|+\left|y-2\right|\le0\)
Mà \(\hept{\begin{cases}\left|x+3\right|\ge0\\\left|y-2\right|\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x+3=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=2\end{cases}}\)
\(\Rightarrow\text{ }\left(x\text{ ; }y\right)=\left(-3\text{ ; }2\right)\)
a) Ta có :
| x + 4 | < 3 <=> -3 < x + 4 < 3
<=> -7 < x < -1
Do x \(\in\)Z nên x = Ơ{ -6 ; -5 ; -4 ; -3 ; -2 }
Vậy x = { -6 ; -5 ; -4 ; -3 ; -2 }
b) Ta có :| x -14 + 17 | + | y + 10 - 12 | \(\le\)0
<=> | x + 3 | + | y - 2 | \(\le\)0 ( *)
Do \(\hept{\begin{cases}\left|x+3\right|+\left|y-2\right|\ge0\\\left|y-2\right|\ge0\end{cases}}\)nên bpt (*) có :
<=> \(\hept{\begin{cases}x+3=0\\y-2=0\end{cases}}\)=> \(\hept{\begin{cases}x=-3\\y=2\end{cases}}\)
Vậy x = -3 ; y = 2