Lời giải:

Đặt $n+1=a^2$ và $2n+1=b^2$ với $a,b$ là số tự nhiên

$\Rightarrow 2a^2-1=b^2$

Nếu $a\vdots 3$ thì $b^2=2a^2-1\equiv -1\equiv 2\pmod 3$ (vô lý do 1 scp không chia 3 dư 2)

$\Rightarrow a$ không chia hết cho 3

$\Rightarrow a^2\equiv 1\pmod 3$

$\Rightarrow n+1\equiv 1\pmod 3$

$\Rightarrow n\equiv 0\pmod 3$ hay $n$ chia hết cho 3 (1)

Mặt khác:

$b^2=2a^2-1$ lẻ nên $b$ lẻ. Đặt $b=2k+1$ với $k$ tự nhiên.

$2a^2-1=(2k+1)^2=4k^2+4k+1$

$\Rightarrow 2a^2=4k^2+4k+2$

$\Rightarrow a^2=2k^2+2k+1$. Do đó $a$ lẻ. Đặt $a=2m+1$ với $m$ tự nhiên.

Khi đđ:

$n+1=(2m+1)^2=4m^2+4m+1\Rightarrow n=4m^2+4m=4m(m+1)$
Hiển nhiên $m(m+1)$ là tích 2 số tự nhiên liên tiếp nên $m(m+1)\vdots 2$

$\Rightarrow n=4m(m+1)\vdots 8(2)$

Từ $(1); (2)\Rightarrow n\vdots 24$. Ta có đpcm.

TH1: p=3

p+2=5; p+4=7

=>Nhận

TH2: p=3k+1

p+2=3k+1+2=3k+3=3(k+1) chia hết cho 3

=>Loại

TH3: p=3k+2

p+4=3k+2+4=3k+6=3(k+2) chia hết cho 3

=>Loại

Vậy: p=3

\(p^5+1782=\left(2x-5\right)^2\)

=>\(\left(2x-5\right)^2=1782+3^5=2025\)

=>\(\left[{}\begin{matrix}2x-5=45\\2x-5=-45\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\\x=-20\end{matrix}\right.\)

Đáp án C 

C. Rừng lá kim nha bạn

ve 1 :Mưa phùn/giăng đầy ngõ 

\(A=\dfrac{2.3.\left(2^3\right)^6.3^{20}-2^2.5.3^{20}.\left(2^2\right)^8}{7.2^2.2^{16}.\left(3^2\right)^{10}-2.3^2.2^{18}.\left(3^3\right)^6}\)

\(=\dfrac{2.3.2^{18}.3^{20}-5.2^2.3^{20}.2^{16}}{7.2^2.2^{16}.3^{20}-2.3^2.2^{18}.3^{18}}=\dfrac{2^{19}.3^{21}-5.2^{18}.3^{20}}{7.2^{18}.3^{20}-2^{20}.3^{20}}\)

\(=\dfrac{2^{18}.3^{20}\left(2.3-5\right)}{2^{18}.3^{20}.\left(7-2^2\right)}=\dfrac{2.3-5}{7-4}=\dfrac{1}{3}\)

Đề hỏi gì em nhỉ?

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Em cần làm gì với các dữ liệu này?

chắc p không phải là số chẵn

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@linhngocnguyen

1 can

2 soon

3 Were

4 Unless

5 If

thi cuối kì lớp 6 toàn trà sữa nhỉ :)))

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