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a, (x+3)(y+2) = 1
=> (x+3) \(\in\)Ư(1) = \(\left\{-1;1\right\}\)
Do (x+3)(y+2) là số dương
=> (x+3) và (y+2) cùng dấu
\(\Rightarrow\hept{\begin{cases}x+3=1\\y+2=1\end{cases}}\)hay \(\hept{\begin{cases}x+3=-1\\y+2=-1\end{cases}}\)
TH1:
\(\hept{\begin{cases}x+3=1\\y+2=1\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)
TH2:
\(\hept{\begin{cases}x+3=-1\\y+2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}}\)
Vậy ............
b, (2x - 5)(y-6) = 17
=> \(\left(2x-5\right)\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
Ta có bảng sau:
| 2x - 5 | -17 | -1 | 1 | 17 |
| x | -6 | 2 | 3 | 11 |
| y - 6 | -1 | -17 | 17 | 1 |
| y | 5 | -11 | 23 | 7 |
Vậy \(\left(x,y\right)\in\left\{\left(-6,5\right);\left(2,-11\right);\left(3,23\right);\left(11,7\right)\right\}\)
c, Tương tự câu b
a,A=|x-7|+12
Vì \(\left|x-7\right|\ge0\forall x\)nên \(\left|x-7\right|+12\ge12\forall x\)
Ta thấy A=12 khi |x-7| = 0 => x-7 = 0 => x = 7
Vậy GTNN của A là 12 khi x = 7
b,B=|x+12|+|y-1|+4
Vì \(\left|x+12\right|\ge0\forall x\)
\(\left|y-1\right|\ge0\forall y\)
nên \(\left|x+12\right|+\left|y-1\right|\ge0\forall x,y\)
\(\Rightarrow\left|x+12\right|+\left|y-1\right|+4\ge4\forall x,y\)
Ta thấy B = 4 khi \(\hept{\begin{cases}\left|x+12\right|=0\\\left|y-1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x+12=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-12\\y=1\end{cases}}\)
Vậy GTNN của B là 4 khi x = -12 và y = 1
a. \(158-x=-12\)
\(x=158+12\)
\(x=170\)
b. \(37+x=12\)
\(x=12-37\)
\(x=-25\)
c. \(2x-15=-47\)
\(2x=\left(-47\right)+15\)
\(x=\dfrac{\left(-32\right)}{2}\)
\(x=-16\)
d. \(\left(-5\right)^2-\left(5x-3\right)=43\)
\(25-\left(5x-3\right)=43\)
\(\left(5x-3\right)=25-43\)
\(5x=\left(-18\right)+3\)
\(x=\dfrac{\left(-15\right)}{5}=-3\)
e. \(\left|x-1\right|+\left(-5\right)=2\)
\(\left|x-1\right|=2-\left(-5\right)\)
\(\left|x-1\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)
f. \(\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g. \(\left|x\right|=12\)
\(\Rightarrow x=\pm12\)
Câu h ko có x sao tìm
a) \(158-x=-12.\)
\(\Rightarrow x=158-\left(-12\right).\)
\(\Rightarrow x=158+12=170.\)
Vậy..........
b) \(37+x=12.\)
\(\Rightarrow x=12-37.\)
\(\Rightarrow x=-25.\)
Vậy..........
c) \(2x-15=-47.\)
\(\Rightarrow2x=-47+15.\)
\(\Rightarrow2x=-32.\)
\(\Rightarrow x=-\dfrac{32}{2}=-16.\)
Vậy..........
d) \(\left(-5\right)^2-\left(5x-3\right)=43.\)
\(\Rightarrow25-\left(5x-3\right)=43.\)
\(\Rightarrow5x-3=25-43.\)
\(\Rightarrow5x-3=-18.\)
\(\Rightarrow5x=-18+3.\)
\(\Rightarrow5x=-15.\)
\(\Rightarrow x=-\dfrac{15}{5}=3.\)
Vậy..........
e) \(\left|x-1\right|+\left(-5\right)=2.\)
\(\Rightarrow\left|x-1\right|=2-\left(-5\right).\)
\(\Rightarrow\left|x-1\right|=2+5.\)
\(\Rightarrow\left|x-1\right|=7.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right..\)
Vậy..........
f) \(\left|x+1\right|=4\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right..\)
Vậy..........
g) \(\left|x\right|=12\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right..\)
Vậy..........
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
a/ -4 là B( a- 1 )
Hay a - 1 là Ư(-4)
Ta có Ư(-4) = { -4; -2; -1; 1; 2 ; 4 }
Xét :
a - 1 = -4 => a = -3
a - 1 = -2 => a = -1
a - 1 = -1 => a = 0
a - 1 = 1 => a= 2
a - 1 = 2 => a = 3
a - 1 = 4 => a = 5