0

Ta có: a = 3k + 1 (k thuộc N)

          b = 3q + 2 (q thuộc N)

=> ab = (3k + 1)(3q + 2) = 9kq + 3q + 6k + 2 = 3(3kq + q + 2k) + 2 chia 3 dư 2 (đpcm)

.............\(\Leftrightarrow px^2+2x-3-qx+3x^2-1\)\(\Leftrightarrow\)\(x^2\left(p+3\right)+x\left(2-q\right)-4=0\)\(\Leftrightarrow\left\{\begin{matrix}p+3=0\\2-q=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}p=-3\\q=2\end{matrix}\right.\)

K=25

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