K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
28 tháng 12 2017
Ta có: \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Rightarrow P=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{5^{32}-1}{2}\)
PT
1
CM
8 tháng 11 2018
Ta có:
( 5 2 - 1).P = ( 5 2 – 1).12.( 5 2 + 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 2 – 1).( 5 2 + 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 4 - 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 8 - 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 16 - 1)( 5 16 + 1)
= 12.( 5 32 - 1)
31 tháng 8 2021
\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\left(5^{128}-1\right)=2.5^{128}-2\)
31 tháng 8 2021
c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{128}-1\right)\)
\(=2\cdot5^{128}-2\)
S
6 tháng 7 2016
P=2.(5^2-1).(5^2+1).(5^4+1).(5^8+1).(5^16+1)
=2.(5^4-1).(5^4+1).(5^8+1).(5^16+1)
= 2.(5^8-1).(5^8+1).(5^16+1)
= 2.(5^16-1).(5^16+1)
= 2.(5^32-1)
25 tháng 7 2016
1)P= 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=> 2P = 24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
=> P = (5^32-1)/2
A
8 tháng 7 2019
\(2\left(x-2\right)\left(x+3\right)-x^2+4=0\)
\(2\left(x^2+3x-2x-6\right)-x^2+4=0\)
\(2x^2+6x-4x-12-x^2+4=0\)
\(x^2+2x-8=0\)
\(x^2+4x-2x-8=0\)
\(x\left(x+4\right)-2\left(x+4\right)=0\)
\(\left(x+4\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+4=0\rightarrow x=\left(-4\right)\\x-2=0\rightarrow x=2\end{cases}}\)
A
8 tháng 7 2019
3/
a/ \(2\left(x+1\right)^2-3\left(x-1\right)^2+\left(x+2\right)\left(5-x\right)\)
\(=2\left(x^2+2x+1\right)-3\left(x^2-2x+1\right)+\left(5x-x^2+10-2x\right)\)
\(=2x^2+4x+2-3x^2+6x-3+5x-x^2+10-2x\)
\(=-2x^2+13x+9\)
b/ \(\left(3x-1\right)^3+\left(3x-1\right)^3-6x^2+9\)
\(=2\left(3x-1\right)^3-6x^2+9\)
\(=2\left(\left(3x\right)^3-3\left(3x\right)^2\cdot1+3\cdot3x\cdot1-1\right)-6x^2+9\)
\(=2\left(27x^3-27x^2+9x-1\right)-6x^2+9\)
\(=54x^3-54x^2+18x-2-6x^2+9\)
\(=54x^3-60x^2+18x+7\)
Số hơi dài, nên dễ tính sai -,- tính mik hay cẩu thả có j sai ibbb ạ
20 tháng 10 2022
Bài4:
=>x(x^2+1)=0
>x=0
Bài 5:
=>\(3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)
PN
19 tháng 12 2015
Công thức tổng quát:
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Do đó:
\(A=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x-4}+\frac{1}{\left(x-1\right)\left(x+10\right)}\)
Bạn tự làm tiếp nhé.
2P = 24.(5^2 + 1 )(5^4 + 1) ... (5^16 + 1)
2P = (5^2 - 1) (5^2 + 1) (5^4 + 1) .. (5^16+1)
2P = (5^4 - 1 )(5^4 + 1 ) (5^8 + 1)
2P = (5^8 - 1 ) (5^8 + 1) (5^16 + 1)
2P = ( 5^ 16 - 1 ) 5^ 16 + 1)
2P = 5^32 - 1
P = (5^32 - 1) : 2
\(P=12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Rightarrow2P=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=5^{32}-1\)
\(\Leftrightarrow P=\frac{5^{32}-1}{2}\)