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1)\(y=\frac{x^2+3x+7}{x+3}=\frac{x\left(x+3\right)+7}{x+3}=x+\frac{7}{x+3}\)= > x +3 thuoc\(U_{\left(7\right)}=\left\{1;-1;7;-7\right\}\)
x thuoc \(\left\{-2;-4;3;-11\right\}\)
2)\(y=\frac{4x+3}{2x+6}=\frac{4x+12-8}{2x+6}=\frac{2\left(2x+6\right)-8}{2x+6}=2-\frac{8}{2x+6}\) =>2x+6 thuoc
\(U_{\left(8\right)}=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>x thuoc \(\left\{-2;-4;-1;-5;1;-7\right\}\)
4)\(y=\frac{4x+1}{3x-1}\)
\(3y=\frac{12x+3}{3x-1}=\frac{12x-4+7}{3x-1}=\frac{4\left(3x-1\right)+7}{3x-1}=4+\frac{7}{3x-1}\)
3x+1 thuoc {1;-1;7;-7}
3x thuoc {0;-2;6;-8}
x thuoc {0;2}
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 , |2x+4|+|y-6|=0
=> 2 x + 4 = 0 => x = 0
=> y - 6 = 0 => y = 6
Vậy x = 0 và y = 6
\(\frac{1}{x}+\frac{y}{3}=\frac{1}{6}\)
=> \(\frac{1}{x}=\frac{1}{6}-\frac{y}{3}\)
=> \(\frac{1}{x}=\frac{1-2y}{6}\)
=> \(x\left(1-2y\right)=6\)
=> \(x;1-2y\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)
Vì \(y\in N\Rightarrow1-2y\in\left\{1;3\right\}\)
\(\Rightarrow x\in\left\{2;6\right\}\)
Lập bảng :
| 1 - 2y | 1 | 3 |
| x | 6 | 2 |
| y | 0 | -1 (loại) |
Vậy ...
\(\frac{3}{4}-2.\left|2x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow2.\left|2x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\Rightarrow2.\left|2x-\frac{2}{3}\right|=\frac{1}{4}\)
\(\Rightarrow\left|2x-\frac{2}{3}\right|=\frac{1}{4}:2\)
\(\Rightarrow\left|2x-\frac{2}{3}\right|=\frac{1}{8}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{2}{3}=\frac{1}{8}\\2x-\frac{2}{3}=\frac{-1}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{1}{8}+\frac{2}{3}\\2x=\frac{-1}{8}+\frac{2}{3}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{19}{24}\\2x=\frac{13}{24}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{19}{24}:2\\x=\frac{13}{24}:2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{19}{48}\\x=\frac{13}{48}\end{cases}}\)
Vậy ...................................
~ Hok tốt ~
a) ( x + 1 ) . ( y + 2 ) = - 5 = -1 .5 = 1.-5 = 5.-1=-5.1
Vậy có 4 cặp (x;y) là: ...
b) ; c) tương tự nhé!
Vậy có 4 cặp số ( x ; y) là: (-2;3) , ( 0;7) , ( 4;-3) , ( -6;-1)