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a) Ta có:
(5^2n+1) + (2^n+4) + (2^n+1) = (25^n).5 - 5.(2^n) + (2^n).( 5 + 2^4 +2) = 5.( 25^n - 2^n ) + 23.2^n chia hết cho 23.
Lời giải:
a)
\(5^{2n+1}+2^{n+4}+2^{n+1}=5.25^n+16.2^n+2.2^n\)
\(\equiv 5.2^n+16.2^n+2.2^n\pmod {23}\)
\(\equiv 23.2^n\equiv 0\pmod {23}\)
Ta có đpcm.
b)
\(2^{2n+2}+24n+14\) hiển nhiên chia hết cho $2(1)$
Mặt khác:
Nếu $n=3k+1$:
$2^{2n+2}+24n+14=2^{6k+4}+72k+38$
$=16.2^{6k}+72k+38\equiv 16+72k+38=54+72k\equiv 0\pmod 9$
Nếu $n=3k$:
$2^{2n+2}+24n+14=2^{6k+2}+72k+14=4.2^{6k}+72k+14$
$\equiv 4+72k+14=18+72k\equiv 0\pmod 9$
Nếu $n=3k+2$:
$2^{2n+2}+24n+14=2^{6k+6}+72k+62\equiv 1+72k+62$
$\equiv 63+72k\equiv 0\pmod 9$
Vậy tóm lại $2^{2n+2}+24n+14$ chia hết cho $9$ (2)
Từ $(1);(2)\Rightarrow 2^{2n+2}+24n+14\vdots 18$ (đpcm)
(3x2 - 51)2n = 576n
=> (3x2 - 51)2n = 242n = (-24)2n
\(\Rightarrow\left[\begin{array}{nghiempt}3x^2-51=24\\3x^2-51=-24\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x^2=75\\3x^2=27\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x^2=25\\x^2=9\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\\\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=5\\x=-5\\x=3\\x=-3\end{array}\right.\)