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a)\(\left(2+1\right)\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
Tiếp tục như thế, ta được:
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)-1=2^{512}-1-1=2^{512}-2\)
b) \(24\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
Tiếp tục như thế, ta được:
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}=5^{64}-1-5^{64}=-1\)
\(\left(2+1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(\left(2-1\right).\left(2+1\right).\left(2^2+1\right).....\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right).\left(2^{256}+1\right)+1=2^{512}+1\)
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
\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)
c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)
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,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)
\(B=\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{32}-1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{64}-1\right).\frac{1}{24}\)
\(B=\frac{5^{64}-1}{24}\)
3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
4:
D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)
=(4^8-1)(4^8+1)*...*(4^64+1)
=...
=4^128-1
5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)
=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1
=5^256-1+5^256-1
=2*5^256-2
thsu là rất ngưỡng mộ anh ạ 🥹 em mấy lần off vì quá nhác nhưng lần nào ngoi lại lên cũng thấy anh cày chăm chỉ quá tr 😭