Từ đầu bài 

=> 1.\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\) \(+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)

=> \(\left(a-b\right)\left(a+b\right)+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}+b^{64}\)

=> \(\left(a^2-b^2\right)\left(a^2+b^2\right)+...+\left(a^{32}+b^{32}\right)\)= a^64 + b^64

tương tự sẽ ra kết quả cuối là \(\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\left(đpcm\right)\)

ta có \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) => \(\frac{a^2-b^2}{a-b}=a+b\)

        \(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)\)=> \(\frac{a^4-b^4}{a^2-b^2}=a^2+b^2\) 

        \(a^8-b^8=\left(a^4-b^4\right)\left(a^4+b^4\right)\) => \(\frac{a^8-b^8}{a^4-b^4}=a^4+b^4\)

        ...............................................................................................

        \(a^{64}-b^{64}=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\) => \(\frac{a^{64}-b^{64}}{a^{32}-b^{32}}=a^{32}+b^{32}\)

thay vào ta được 

\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)......\left(a^{32}+b^{32}\right)\)

\(=\frac{a^2-b^2}{a-b}.\frac{a^4-b^4}{a^2-b^2}.\frac{a^8-b^8}{a^4-b^4}.............\frac{a ^{64}-b^{64}}{a^{32}-b^{32}}\)

\(=\frac{a^{64}-b^{64}}{a-b}\)

mà a-b= 1 nên \(\frac{a^{64}-b^{64}}{a-b}=a^{64}-b^{64}\)

a) (2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^4-1)(2^4+1)....(2^32+1)-2^64

=......

=(2^32-1)(2^32+1)-2^64

=2^64-1-2^64=-1

b)Đặt A=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)+(5^128-3^128)/2

đặt B=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)

\(2B=\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^4-3^4\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=.......\)

2B=(5^64-3^64)(5^64+3^64)

2B=5^128-3^128

B=(5^128-3^128)/2 (thế vào đề bài)

=> A=B+(5^128-3^128)/2=(5^128-3^128)/2+(5^128-3^128)/2=\(\frac{2\left(5^{128}-3^{128}\right)}{2}=\left(5^{128}-3^{128}\right)\)

a) A = ( 2-1)(2+1)(2²+1)...(2³²+1)-2⁶⁴

         =(2²-1)(2²+1)(2⁴+1)... -2⁶⁴

       =....

       =2⁶⁴-1-2⁶⁴=1

câu b tương tự nhá

a, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=2^{64}-1-2^{64}=-1\)

b,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)

Có a = b+1

=> a - b =1

=> (a-b)(a+b)(a^2+b^2)(a^4+b^4)...(a^32+b^32) = (a-b)(a^64-b^64)

=> (a^2-b^2)(a^2+b^2)(a^4+b^4)...(a^32+b^32) = 1 . (a^64 - b^64)

=> (a^4-b^4)(a^4+b^4)(a^8+b^8)(a^16+b^16)(a^32+b^32) = a^64 - b^64

=> (a^8-b^8)(a^8+b^8)(a^16+b^16)(a^32+b^32) = a^64 - b^64

=> (a^16-b^16)(a^16+b^16)(a^32+b^32) = a^64 - b^64

=> (a^32-b^32)(a^32+b^32) = a^64 - b^64

=> a^64-b^64 = a^64 - b^64

=> đpcm

Sao tự nhiên lại lòi ra số c vậy?

\(A=\left(6+1\right)\left(6^2+1\right)\cdot...\cdot\left(6^{64}+1\right)\)

\(\left(6-1\right)A=\left(6-1\right)\left(6+1\right)\left(6^2+1\right)\cdot...\cdot\left(6^{64}+1\right)\)

\(5A=\left(6^2-1\right)\left(6^2+1\right)\cdot...\cdot\left(6^{64}+1\right)\)

\(5A=\left(6^{64}-1\right)\left(6^{64}+1\right)\)

\(5A=6^{128}-1=B\)

\(\Rightarrow B=5A\)

nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)

\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

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

\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)

\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)

g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)

\(=\left(2a+2b\right)^2-2\left(a+b+c+d\right)\left(a+b-c-d\right)+\left(2a-2b\right)^2-2\left(a+c-b-d\right)\left(a+d-b-c\right)\)

\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)

\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)

\(=2\left(\left(2a\right)^2-2\left(a+b\right)\left(a-b\right)\right)+2\left(\left(2c\right)^2-2\left(c+d\right)\left(c-d\right)\right)\)

\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)

\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)

bn đăng nhiều quá nên mk làm câu nào hay câu đó nha

mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .