\(C=\frac{15x^2-7x-5}{2x-3}=\frac{7x\left(2x-3\right)+7\left(2x-3\right)+\left(21-5+x^2\right)}{2x-3}=7x+7+\left(\frac{16+x^2}{2x-3}\right)\)
x nguyên =>\(\frac{x^2+16}{2x-3}=a\in Z\) và 2x-3 khác 0 (*)
\(\Leftrightarrow x^2+16=\left(2x-3\right)a\Leftrightarrow x^2-2ax+16=-3a\)
\(\Leftrightarrow\left(x^2-2ax+a^2\right)=a^2-3a-16\)
\(\Leftrightarrow\left(x-a\right)^2=a^2-3a-16\) (**)
VT là số CP => đk cần VP là số cp
\(a^2-3a-16=k^2\Leftrightarrow\left(2a-3\right)^2-73=4k^2\Leftrightarrow t^2-\left(2k\right)^2=73\)
Hệ nghiệm nguyên =>nghiệm duy nhất !2t!=74=> !t!=37=> !2a-3!=37
=>\(\left[\begin{matrix}a=20\\a=-17\end{matrix}\right.\)
(*)&(**)
\(\left[\begin{matrix}\left(x-20\right)^2=324=18^{2\left(1\right)}\\\left(x+17\right)^2=324=18^2\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[\begin{matrix}x-20=-18\\x-20=18\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=2\\x=48\end{matrix}\right.\) \(\left(2\right)\Rightarrow\left[\begin{matrix}x=1\\x=-35\end{matrix}\right.\)
Đáp số: x={-35,1,2,48}
