b: \(A=\left(a-1\right)\left(b-1\right)\)
\(=\left[\left(2k+1\right)^2-1\right]\left[\left(2k+3\right)^2-1\right]\)
\(=\left[4k^2+4k\right]\left(2k+4\right)\left(2k+2\right)\)
\(=4k\left(k+1\right)\cdot4\left(k+1\right)\left(k+2\right)\)
\(=16k\left(k+1\right)^2\cdot\left(k+2\right)\)
Vì k(k+1) chia hết cho 2 nên A chia hết cho 2
Vì (k+1)(k+2) chia hết cho 2 nên k(k+1)(k+2)(k+1) chia hết cho 4
Vì k(k+1)(k+2) chia hết cho 3 nên A chia hết cho 3
=>k(k+1)2(k+2) chia hết cho 12
=>A chia hết cho 192
a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x-6+6}{x-2}-\dfrac{2y+2-2}{y+1}=8\\\dfrac{x-2+2}{x-2}+\dfrac{3y+3-3}{y+1}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-2}+\dfrac{2}{y+1}=8-3+2=7\\\dfrac{2}{x-2}+\dfrac{-3}{y+1}=-1-1-3=-5\end{matrix}\right.\)
=>x-2=2; y+1=1/2
=>x=4; y=-1/2
