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

\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)

\(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^2\right)\left(1+3^2\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^4\right)\left(1+3^4\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^8\right)\left(1+3^8\right)\left(3^{16}+1\right)\\ =\left(1-3^{16}\right)\left(1+3^{16}\right)=1-3^{32}\)

a) \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)

\(=\left(a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc\right)-\left(b^2-2bc+c^2\right)+2ab-2ac\)

\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)

\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)

\(=a^2\)

Nếu đề thế này thì mình có thể làm được:

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=3^{32}-1\)

\(\Rightarrow A=\dfrac{3^{32}-1}{2}\)

=> B>A

B>A

Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)

Ta có: B=\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow\) 2B= \(2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3-1\right).\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^{16}-1\right)\left(3^{16}+1\right)\)

= \(3^{32}-1\)

\(\Rightarrow\) B= \(\dfrac{3^{32}-1}{2}\)

Mà ta có A= \(3^{32}-1\)

\(\Rightarrow\) A=2B

\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)+11\)

\(=\left(3^{64}-1\right)+11=3^{64}+10\)

A = 8.(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1) + 1

A = (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1) + 1

A = (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1) + 1

A = (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1) + 1

A = (3¹⁶ - 1)(3¹⁶ + 1)(3³² + 1) + 1

A = (3³² - 1)(3³² + 1) + 1

A = 3⁶⁴ - 1 + 1

A = 3⁶⁴

Lời giải :

\(\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{4}\cdot\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{4}\cdot\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{4}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{4}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{4}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\frac{1}{4}\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(=\frac{3^{64}-1}{4}\)

Thank you anh

[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

Ta có: \(A=3^{32}-1=\left(3^{16}+1\right)\left(3^{16}-1\right)\)

\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^8-1\right)\)

\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)

\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)\)

\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)

\(=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

Vậy A = 2B