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đáp án
công thức tổng quát là \(\frac{x}{2^x}\)cho x chạy từ 1-10 bằng xích ma là ra 509/256
![](https://rs.olm.vn/images/avt/0.png?1311)
TUI ĐANG GẤP CHO TÔI HỎI BÀI NÀY LỚP 2 NHA\\\\
AN CÓ 180 CÁI KẸO.BÌNH CÓ 160. HỎI BÌNH CÓ MẤY CÁI KẸO
a) Ta có: \(2.4.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)
=(3^8-1)(3^8+1)(3^16+1)
=(3^16-1)(3^16+1)
=3^32-1
2: B=(1-3^2)(1+3^2)*...*(1+3^16)
=(1-3^4)(1+3^4)(1+3^8)(1+3^16)
=1-3^32
1
\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
\(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^2\right)\left(1+3^2\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^4\right)\left(1+3^4\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^8\right)\left(1+3^8\right)\left(3^{16}+1\right)\\ =\left(1-3^{16}\right)\left(1+3^{16}\right)=1-3^{32}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+3+3^2+3^3+3^4\)
\(A=1+3+9+27+81\)
\(A=121\)
\(B=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(8B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(8B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(8B=\left(3^8-1\right)\left(3^8+1\right)\)
\(8B=3^{16}-1\)
\(B=\frac{1}{8}\left(3^{16}-1\right)\)
\(B=\frac{3^{16}}{8}-\frac{1}{8}\)
\(B=\text{ }\frac{43046721}{8}-\frac{1}{8}\)
\(B=\frac{43046720}{8}\)
\(B=5380840\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+3+3^2+3^3+3^4
=1+3+9+27+81
=(9+81)+(3+27)+1
=90+30+1
=121
B=(3^2+1)×(3^4+1)×(3^8+1)
=(9+1)×(81+1)×(6561+1)
=10x82x6562
=820x6562
=5380840
A =1+3+32+33+34
3A =3+32+33+35
3A-A=35-1
2A =243-1
A=242:2
A=121
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
Nhân cả hai vế với 3-1, ta được:
(3-1)A=(3+1)(3-1)(3^2+1)...(3^2048+1)
(3-1)A=(3^2-1)(3^2+1)....(3^2048+1)
........
(3-1)A=(3^2048-1)(3^2048+1)
(3-1)A=3^4096-1=> A=\(\frac{3^{4096}-1}{3-1}\)
(Nếu có thể thu gọn tiếp thì bạn cứ thế mà tính)