Bài 48: Tìm số nguyên x, biết:
a) | x + 2| = 0 b) | x - 5| = |-7|
c) | x - 3 | = 7 - ( -2) d) ( 7 - x) - ( 25 + 7 ) = - 25
e) | x - 3| = |5| + | -7| g) 4 - ( 7 - x) = x - ( 13 -4) :
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a)
\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng:
x+1 | 1 | -1 | 5 | -5 |
y-2 | 5 | -5 | 1 | -1 |
x | 0 | -2 | 4 | -6 |
y | 7 | -3 | 3 | 1 |
Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)
b)
\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Ta có bảng:
x-5 | 1 | -1 | 7 | -7 |
y+4 | -7 | 7 | -1 | 1 |
x | 6 | 4 | 12 | -2 |
y | -11 | 3 | -5 | -3 |
Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)
b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-7;-9;-3;-13\right\}\)
B3 a) x=4 b) x=-7 c) x=5 d) x=4
B2 a) -3+ -2+ -1+0+1+2+3+4=4
b) -6+ -5+ -4+ -3+ -2+ -1+0+1+2+3+4=-11
c) -18+-17+-16+-15+-14+-13+-12+-11+-10+-9+-8+-7+-6+-5+-4+3+-2+-1+0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19=19
Bài 1:
a)-54
b)-8
Bài 2:
a)(x-14):5=415:413
⇔(x-14):5=42
⇔(x-14):5=16
⇔x-14=80
⇔x=94
b)7x-15x=15-175
⇔-8x=-160
⇔x=20
a) Ta có: 12-5x=37
\(\Leftrightarrow5x=-25\)
hay x=-5
Vậy: x=-5
b) Ta có: 7-3|x-2|=-11
\(\Leftrightarrow3\left|x-2\right|=18\)
\(\Leftrightarrow\left|x-2\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-4\right\}\)
c) Ta có: \(x+\dfrac{2}{8}=-\dfrac{15}{4}\)
\(\Leftrightarrow x=\dfrac{-15}{4}-\dfrac{2}{8}=\dfrac{-15}{4}-\dfrac{1}{4}\)
hay x=-4
Vậy: x=-4
a, \(\Leftrightarrow5x=12-37=-25\)
\(\Leftrightarrow x=-\dfrac{25}{5}=-5\)
Vậy ...
b, \(\Leftrightarrow3\left|x-2\right|=7+11=18\)
\(\Leftrightarrow\left|x-2\right|=\dfrac{18}{3}=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
Vậy ...
c, \(\Leftrightarrow x=-\dfrac{15}{4}-\dfrac{2}{8}=-4\)
Vậy ..
a) | x + 2| = 0
=>x+2=0
x=0-2
x=-2
Vậy x=-2
b) | x - 5| = |-7|
|x-5|=7
\(\Rightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7+5=12\\x=-12+5=-7\end{cases}}\)
Vậy..............................................
c) | x - 3 | = 7 - ( -2)
| x-3|=9
* x-3=9 *x-3=-9
x=9+3 x=-9+3
x=12 x=-6
Vậy........................................
d) ( 7 - x) - ( 25 + 7 ) = - 25
(7-x)-32=-25
7-x=-25+32
7-x=7
x=7-7
x=0
Vậy x=0
e) | x - 3| = |5| + | -7|
|x-3|=12
* x-3=12 * x-3=-12
x=12+3 x=-12+3
x= 15 x=-9
Vậy...............................................
g) 4 - ( 7 - x) = x - ( 13 -4)
4-7-x=x-9
3-x=x-9
3+9=x+x
12=2x
x=12:2
x=6
Vậy x=6
\(a,|x+2|=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
\(b,|x-5|=|-7|\)
\(\Leftrightarrow|x-5|=7\Leftrightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}\Rightarrow\orbr{\begin{cases}x=12\\x=-2\end{cases}}}\)
\(c,|x-3|=7-\left(-2\right)\)
\(|x-3|=9\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)
\(d,\left(7-x\right)-\left(25+7\right)=-25\)
\(\left(7-x\right)-32=-25\)
\(7-x=7\Leftrightarrow x=0\)
\(e,|x-3|=|5|+|-7|\)
\(|x-3|=12\Leftrightarrow\orbr{\begin{cases}x-3=12\\x-3=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=15\\x=-9\end{cases}}}\)
\(g,4-\left(7-x\right)=x-\left(13-4\right)\)
\(4-7+x=x-13+4\)
\(-3+x=x-17\)
\(x-x=-17+3\)
\(-2x=-14\)ko chắc chỗ này vì -x-x=-2x nha bn
\(x=7\)