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A = \(\frac{1}{2}.24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
A = \(\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
A = \(\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
A = \(\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
A = \(\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
A = \(\frac{1}{2}\left(5^{32}-1\right)\left(5^{32}+1\right)\)
A = \(\frac{1}{2}\left(5^{64}-1\right)\)
\(2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(=5^{64}-1\)
=> \(A=\frac{5^{64}-1}{2}\)
( bài này áp dụng hằng đẳng thức \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
Ta có
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(=2^{64}-1\)
3.(22+1)(24+1)(28+1)(216+1)(232+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)
=(24-1)(24+1)(28+1)(216+1)(232+1)
=(28-1)(28+1)(216+1)(232+1)
=(216-1)(216+1)(232+1)
=(232-1)(232+1)
=264-1
3(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(24-1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(28-1)(28+1)(216+1)(232+1)(264+1)
=(216-1)(216+1)(232+1)(264+1)
=(232-1)(232+1)(264+1)
=(264-1)(264+1)
=(2128-1)
Nếu thấy đúng thì thích cho mình nha
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(=2^{64}-1\)
A = 3( 22 + 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 22 - 1 )( 22 + 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 24 - 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 28 - 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 216 - 1 )( 216 + 1 )( 232 + 1 )
= ( 232 - 1 )( 232 + 1 )
= 264 - 1
a) Sai đề nên sửa luôn\(\left(2x-5\right)\left(4x^2+10x+25\right)-2x\left(2x+1\right)^2+8x^2+23x+125\)
=\(8x^3-125-2x\left(4x^2+4x+1\right)+8x^2+23x+125\)
= \(8x^3-125-8x^3-8x^2-2x+8x^2+23x+125\)
= \(21x\)
b) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^{16}-1\right)\left(2^{16}+1\right)-2^{32}\)
= \(2^{32}-1-2^{32}=-1\)
A=(2+1)x(22+1)x(24+1)x(28+1)x(216+1)
= 3.5.17.257.65537
b) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{64}-1\right)-2^{64}\)
\(=-1\)
\(\left(1^2-2^2\right)+\left(3^2-4^2\right)+....+\left(99^2-100^2\right)\)
\(=\left(1-2\right)\left(2+1\right)+\left(3-4\right)\left(4+3\right)+....+\left(99-100\right)\left(100+99\right)\)
\(=\left(-1\right)\left(1+2+3+....+100\right)=\frac{\left(-1\right)100.99}{2}=-4950\)
a) \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2\)
\(=16\)
b) \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{64}-1\right)\)
Ta có: (2+1)(22+1)(24+1)(28+1)(216+1)(232+1)
= (2-1)(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)
= (22-1)(22+1)(24+1)(28+1)(216+1)(232+1)
= (24-1)(24+1)(28+1)(216+1)(232+1)
= (28-1)(28+1)(216+1)(232+1)
= (216-1)(216+1)(232+1)
= (232-1)(232-1)
= 264-1
đặt biểu thức là A. Xong rùi đó, A đơn giản nhất còn gì