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14 tháng 8 2019

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(P=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(P=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(P=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(P=\frac{1}{2}\left(5^{32}+1\right)\)

19 tháng 8 2020

Đặt \(A=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

\(\Rightarrow2A=24.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^2-1\right).\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^4-1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^8-1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^{16}-1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^{16}\right)^2-1^2\)

     \(2A=5^{32}-1\)

\(\Rightarrow A=\frac{5^{32}-1}{2}.\)

8 tháng 7 2019

3. ( 22 + 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1

= ( 22 - 1 ).( 22 + 1 ).( 24 + 1 ).( 28 + 1 )....( 264 + 1 ) + 1

= ( 24 - 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1

= ( 28 + 1 ).....( 264 + 1 )  + 1

= ( 264 - 1 ).( 264 + 1 ) + 1

=  2128 - 1 + 1

= 2128

8 tháng 7 2019

8.( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 32 - 1 ).( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 34 - 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 38 - 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 316 - 1 )......( 3128 + 1 ) + 1

= ( 3128 - 1 ).( 3128 + 1 ) + 1

=  3256 - 1 + 1

= 3256

11 tháng 8 2016

 nghe nhe',bai nay de thui ma. 
ta xet ve trai a^3+b^3+c^3= 
[(a+b)(a^2-ab+b^2)]+c^3 dung ko.(1) 
ma ta co theo gia thiet a+b+c=0 suy ra c= - (a+b)suy ra 
c^3= -(a+b)^3 
thay vao`(1) ta co [(a+b)(a^2-ab+b^2)] - (a+b)^3 
(lay nhan tu chung ta co)=(a+b)[a^2-ab+b^2-(a+b)^2] 
(phan h (a+b)^2) =(a+b)[a^2-ab+b^2-(a^2+2ab+b^2)] 
=(a+b)(a^2-ab+b^2-a^2-2ab-b^2) 
=(a+b).(-3ab) 
= -(a+b).3ab (2) 
theo gia thiet ta co a+b+c=0 suy ra c= -(a+b) 
thay vao(2) ta dc 
=3abc 
vay la xong 
ket luan ve trai bang ve phai 

k cho mk nha

11 tháng 8 2016

Mơn bạn 

đã đúng 

17 tháng 7 2016

A = 4( x+2x.1/4+1/16) +1

GTNN  A = 3/4

26 tháng 6 2016

\(p=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^{32}-1\right)=\frac{5^{32}-1}{2}\)

29 tháng 8 2020

                      Bài làm : 

Ta có:

 \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(P=\frac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{5^{32}-1}{2}\)

\(\text{Vậy : }P=\frac{5^{32}-1}{2}\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

29 tháng 8 2020

Bài làm:

Đặt \(A=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

=> \(2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

<=> \(2A=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

<=> \(2A=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

<=> \(2A=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

<=> \(2A=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

<=> \(2A=5^{32}-1\)

=> \(A=\frac{5^{32}-1}{2}\)