rút gọn biểu thức
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
18 tháng 6 2018
a) \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=\left(a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc\right)-\left(b^2-2bc+c^2\right)+2ab-2ac\)
\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)
\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)
\(=a^2\)
PN
2
18 tháng 12 2016
2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
12 tháng 12 2017
2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
chúc bn hok tốt @_@
6 tháng 10 2019
Áp dụng HĐT đáng nhớ :
\(\left(a-b\right)\left(a+b\right)=a^2-b^2\) . Ta có :
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
Chúc bạn học tốt !!!
13 tháng 10 2019
a) \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-\left(x^2+x-3x-3\right)\)
\(=x^2-4-x^2-x+3x+3\)
\(=2x-1\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
20 tháng 11 2019
3(22 + 1)(24 + 1)(28 + 1)(216 + 1)
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
= (24 - 1)(24 + 1)(28 + 1)(216 + 1)
= (28 - 1)(28 + 1)(216 + 1)
= (216 - 1)(216 + 1)
= 232 - 1
AH
Akai Haruma
Giáo viên
6 tháng 10 2019
Lời giải:
Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:
\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
AH
Akai Haruma
Giáo viên
3 tháng 10 2019
Lời giải:
Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:
\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
[Toán 8] Rút gọn $ (3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)$ | HOCMAI Forum - Cộng đồng học sinh Việt Nam