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Đặt \(4p+1=a^2\)
Dễ thấy 4p+1 lẻ nên a lẻ.Đặt a=2k+1
Khi đó \(4p+1=4k^2+4k+1\)
\(\Rightarrow p=2\left(k+1\right)\)
\(\Rightarrow p=2\) vì \(2\left(k+1\right)\) chẵn
Ta có: \(4p+1=n^2\left(n\inℕ\right)\)
\(4p=n^2-1\)
\(4p=n^2-1^2\)
\(4p=\left(n+1\right)\left(n-1\right)\)
Điều này chỉ xảy ra khi n-1=4 , n+1=p hoặc n+1=4 , n-1=p. Trường hợp thứ nhất cho ta p=6 (không thoả mãn vì 6 là hợp số). Trường hợp thứ hai cho ta p=2 (thoả mãn đề bài). Vậy đáp số của bài toán là p=2.
Có gì ko hiểu hỏi mik nhé!
\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}n+1=4\\n-1=p\end{cases}}\\\hept{\begin{cases}n+1=p\\n-1=4\end{cases}}\end{cases}}\)
a, -1+3 - 5 + 7 - ...... +97 - 99
[ - 1+ 3] - [ 5 + 7] - .... - [ 95 + 97] - 99
[2 - 12] - ..... - [184 - 192] - 99
còn lại tự giải
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
a) 14-(7-x+3)=5-{4-(5- |3| ) }
14-(10-x) = 5-{4-(5-3) }
x +14-10=5-(4-2)
x+4 = 5-2
x+4 =3
x =3-4
x =-1 Vậy x= -1
-7 + [ - (-3) + |6| - (544 + |-6 |) ] = 5 - ( 7 - x + 4)
-7+{ 3+6-(544+6) } =5-(11-x)
-7+(9-600) =x+5-11
-7+-591 =x+(-6)
-598 = x+ (-6)
x =-598 - (-6)
x = -592
Vậy x= -592
tick mình nha
Nếu 2 đường thẳng có 2 điểm chung thì chúng:
A, cắt nhau
B, song song
C, trùng nhau
D, ko xác định
1 - {2 - [3 - 4.(5 - x)]} = 6
2 - [3 - 4.(5 - x)] = 1 - 6 = -5
3 - 4.(5 - x) = 2- (-5) = 7
4.(5 - x) = 3 - 7 = -4
5 - x = (-4) : 4 = -1
x = 5 - (-1)
x = 6
Books have been one of my best friends which have supported me in every step of my life. And the one that I have the deepest impression on is “The miracle of the Namiya general store” .
The book is about three delinquents who were running away from their wrongdoings then accidentally found an old house and hid there for the night. The house turned out to be an abandoned general store where people could seek advice for their troubles by leaving a letter in the mailbox. Miracle happened when the time line somehow switched and letters from 30 years ago were delivered to them. Although none of them ever seriously considered others’ problems, something from the inside urged them to write responses to the troubled people, on behalf of Namiya – the old owner.
“ Miraculous” is exactly how I want to describe this book. No need for dogma lessons, it presents the value of kindness and compassion through different short stories that are linked perfectly together and leaves me hopeful about human nature. The past, present and future are combined flexibly, which creates many a surprise to me. How did the letters change people’s lives? Could the delinquents - whose past was covered by darkness – be awoken and open their hearts to heal the grieving souls? The story presents an open ending but I have got the answer of my own. To any book lovers especially those who have interest in soothing and touching stories, “The miracle of the Namiya general store” by Higashino Keigo is the one that should not be missed.
TƯỞNG GÌ KHÓ , THAM KHẢO NHA BẠN
Đặt \(S=\frac{1}{5}+\frac{2}{5^2}+...+\frac{2010}{5^{2010}}\)
\(\Rightarrow5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2010}{5^{2009}}\)
\(\Rightarrow5S-S=\left(1+...+\frac{2010}{5^{2009}}\right)-\left(\frac{1}{5}+...+\frac{2010}{5^{2010}}\right)\)
\(\Rightarrow4S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2009}}+\frac{2010}{5^{2010}}\)
Đặt \(A=\frac{1}{5}+...+\frac{1}{5^{2009}}\)
\(\Rightarrow5A=1+...+\frac{1}{5^{2008}}\)
\(\Rightarrow5A-A=\left(1+...+\frac{1}{5^{2008}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{2009}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{2009}}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{5^{2009}.4}< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{4}\)
\(\Rightarrow4S< 1+\frac{1}{4}\)
\(\Rightarrow4S< \frac{5}{4}\)
\(\Rightarrow S< \frac{5}{16}\left(đpcm\right)\)
Anh xin lỗi nhé dòng thứ 4 là \(4S=1+\frac{1}{5}+...+\frac{1}{5^{2009}}-\frac{2010}{5^{2010}}\)