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19 tháng 11 2016

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

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

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

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

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

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

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

19 tháng 11 2016

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

11 tháng 7 2018

\(P=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)

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\(T=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(\Rightarrow2T=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(2T=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(2T=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(2T=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)

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

\(\Rightarrow T=\dfrac{3^{16}-1}{2}=21523360\)

bạn ơi bạn viết rõ bài P dc k o mình đọc chả hiểu j

4 tháng 7 2019

Lời giải :

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

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

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

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

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

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

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

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

4 tháng 7 2019

Thank you anh

4 tháng 8 2016

[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

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 !!!

28 tháng 10 2019

1)

a) \(\left(-48\right)^3:16^3\)

\(=\left(-48:16\right)^3\)

\(=\left(-3\right)^3\)

\(=-27.\)

b) \(\left(\frac{9}{10}\right)^6:\left(\frac{17}{-20}\right)^6\)

\(=\left(\frac{9}{10}:\frac{17}{-20}\right)^6\)

\(=\left(-\frac{18}{17}\right)^6\)

Chúc em học tốt!

28 tháng 10 2019

\(\frac{-13^3}{\left(2^3\right)^3}:\frac{\left(-2^5\right)^4}{13^4}\)1.

a, (-48)3:163

= \(\left(\frac{-48}{16}\right)^3\)

= (-3)3

b,\(\left(\frac{9}{10}\right)^6\):\(\left(\frac{17}{-20}\right)^6\)

= \(\left(\frac{9}{10}:\frac{17}{-20}\right)^6\)

=\(\left(\frac{-18}{17}\right)^6\)

c, \(\left(\frac{-13}{8}\right)^3:\left(\frac{-32}{13}\right)^4\)

= \(\frac{-13^3}{\left(2^3\right)^3}:\frac{\left(-2^5\right)^4}{13^4}\)

= \(\frac{-13^3}{2^9}.\frac{-13^4}{2^{20}}\)

=\(\frac{13^7}{2^{29}}\)

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

7 tháng 7 2023

1

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

 

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