\(\dfrac{1}{a+1}+\dfrac{1}{b+1}=\dfrac{1}{2}\left(a,b\ne-1\right)\\ \Rightarrow2\left(a+b+2\right)=\left(a+1\right)\left(b+1\right)\\ \Rightarrow2a+2b+4=ab+a+b+1\\ \Rightarrow a+b-ab+3=0\\ \Rightarrow\left(b-1\right)-a\left(b-1\right)=-4\\ \Rightarrow\left(a-1\right)\left(b-1\right)=4=1\cdot4=2\cdot2\)
| \(a-1\) | 1 | 4 | 2 |
| \(b-1\) | 4 | 1 | 2 |
| \(a\) | 2 | 5 | 3 |
| \(b\) | 5 | 2 | 3 |
Vậy \(\left(a;b\right)=\left(2;5\right);\left(5;2\right);\left(3;3\right)\)
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}=\dfrac{1}{2}\Leftrightarrow\dfrac{2\left(a+1\right)+2\left(b+1\right)-\left(a+1\right)\left(b+1\right)}{2\left(a+b\right)\left(b+1\right)}=0\)
\(\Leftrightarrow a+b-ab+3=0\Leftrightarrow a\left(1-b\right)-\left(1-b\right)=-4\Leftrightarrow\left(a-1\right)\left(1-b\right)=-4\)
Do \(a,b\in N\) nên ta có bảng sau:
| a-1 | -1 | 1 | -4 | 4 | -2 | 2 |
| 1-b | 4 | -4 | 1 | -1 | 2 | -2 |
| a | 0 | 2 | -3(loại) | 5 | -1(loại) | 3 |
| b | -3(loại) | 5 | 0 | 2 | -1(loại) | 3 |
Vậy \(\left(a;b\right)\in\left\{\left(2;5\right);\left(5;2\right);\left(3;3\right)\right\}\)
