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B=(2+1)(22+1)(24+1)(28+1)(216+1)−232
=1.(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(22-1)(22+1)(24+1)(28+1)(216+1)−232
=(24-1)(24+1)(28+1)(216+1)−232
=(28-1)(28+1)(216+1)−232
=(216-1)(216+1)−232
=232-1-232
=-1
A = ( 2 +1 )( 2^2 + 1 )...(2^16+1) - 2^32
A = ( 2 - 1) ( 2 + 1 )(2^2 + 1) .... (2^16 + 1) - 2^32
A = (2^2 - 1) (2^2 + 1) ...(2^16 + 1) - 2^32
A =( 2^ 4 - 1)( 2^4 + 1 )( 2^8 + 1) (2^16+1) -2^32
A = ( 2^8 - 1)( 2^ 8 + 1) ( 2^ 16 + 1)- 2^32
A = ( 2^16 - 1 )( 2^16 + 1) - 2^32
A = 2^32 - 1 - 2^32
A = - 1
dùng hằng đẳng thức A^2 - B^2 = (A - B)(A + B) nhé phần b chuyển vế sang rồi dùng hđt là Okay
Bài 1 :
Ta có : \(VP=\left(a+b\right)^4=\left(a+b\right)\left(a+b\right)^3\)
\(=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
=> HĐT ko đc CM
Bài 2 :
a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)
\(=x^3+2x^2+4x-2x^2-4x-8-x+1+7=x^3-x=x\left(x^2-1\right)\)
Sửa đề : b, \(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=8\left(x^3-1\right)-8x^3+1=8x^3-8-8x^3+1=-7\)
Xin phép chủ nahf cho mjnh sửa đề:D
\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
a,\(\left(a+b\right)^4\)
\(=\left[\left(a+b\right)^2\right]^2\)
\(=\left(a^2+2ab+b^2\right)^2\)
\(=\left[\left(a^2+2ab\right)+b^2\right]^2\)
\(=\left(a^2+2ab\right)^2+2\left(a^2+2ab\right)b^2+b^4\)
\(=a^4+4a^3b+4a^2b^2+2a^2b^2+4ab^3+b^4\)
\(=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
Bài 2:
a,\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)
\(=\left(x^3-8\right)-\left(x-1\right)+7\)
b,\(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x-1\right)\)
\(=8\left(x^3-1\right)-\left(8x^3-1\right)\)
\(=8x^3-8-8x^3+1\)
\(=-7\)
a, Ta co : A = 1999 * 2001
= ( 2000 - 1 ) *( 2000 + 1 )
= \(2000^2-1\)
Vây A < B
cậu ơi tối mình về mình làm tiếp cho bây giờ mình phải đi hok .
( 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
Do : b + 1 = a --> a - b = 1
Ta có : ( a + b)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= 1.( a + b)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a - b)( a + b)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a2 - b2)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a4 - b4)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a8 - b8)( a8 + b8)( a16 + b16)
= ( a16 - b16)( a16 + b16)
= a32 - b32 ( đpcm)
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)