12(52 +1)(54+1)(58 +1)(516 +1)
phép trên thuộc 7 hđt mở rộng nhá (sử dụng hđt 3 bằng phương pháp trung hồi )
làm hộ mình với ............
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11 tháng 8 2019
Mình sửa đề bài nha:
\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{32}-1\right)\)
\(=\frac{5^{32}-1}{2}\)
Chúc bạn học tốt!
S
13 tháng 8 2019
\(4A=4\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5^2-1\right)\left(5^2+1\right).....\left(5^{2048}+1\right)\left(hdt\left(a-b\right)\left(a+b\right)=a^2-b^2\right)=\left(5^4-1\right)\left(5^4+1\right)......\left(5^{2048}+1\right)=\left(5^8-1\right).....\left(5^{2048}+1\right)=.....=\left(5^{1024}+1\right)\left(5^{1024}-1\right)\left(5^{2048}+1\right)=\left(5^{2048}-1\right)\left(5^{2048}+1\right)=5^{4096}-1\)
LT
11 tháng 8 2019
Sửa đề
B = 2(3+1)(32+1)(34+1)(38+1)(316+1)
= (3-1)(3+1)(32+1)(34+1)(38+1)(316+1)
= (32-1)(32+1)(34+1)(38+1)(316+1)
= (34-1)(34+1)(38+1)(316+1)
= (38-1)(38+1)(316+1)
= (316-1)(316+1)
= (332-1)
11 tháng 8 2019
I am a loser: Bạn chép đề sai nha, mình sửa luôn.
\(A=3\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^4-1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^8-1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^{16}-1\right)\cdot\left(2^{16}+1\right)\)
\(A=2^{32}-1\)
Vậy...
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)
MC
10 tháng 8 2019
\(A=3\left(2^3+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right).9\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{32}-1\right)\)
LT
11 tháng 8 2019
12
= \(\frac{24}{2}\)
= \(\frac{1}{2}\left(25-1\right)\)
= \(\frac{1}{2}\left(5^2-1\right)\)
Chép đề sai kìa