Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
a: \(A=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+...+3+2+1\)
=5050
b: \(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\cdot\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(=2^{128}\)
c: \(\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
\(=2c^2\)
1: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x\)
\(=9x\)
2: \(=2\left(a+b\right)^2+2c^2+4a^2-4ab+b^2\)
\(=2a^2+4ab+2b^2+2c^2+4a^2-4ab+b^2\)
\(=6a^2+3b^2+2c^2\)
3: =100+99+98+...+2+1
=5050
4: \(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1=2^{128}\)
\(\left(a+b\right)^3-3ab\left(a+b\right)\\ =\left(a^3+3a^2b+3ab^2+b^3\right)-3a^2b-3ab^2\\ =a^3+b^3\)
b.
\(a^3+b^3+c^3-3abc\\ =\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\\ =\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+2ab+b^2-ab-bc+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Bài 2:
a: \(\left(a-b-2\right)^2-\left(2a-2b\right)\left(a-b-2\right)+a^2-2ab+b^2\)
\(=\left(a-b\right)^2-4\left(a-b\right)+4+\left(a-b\right)^2-2\left(a-b\right)\left(a-b-2\right)\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left[\left(a-b\right)^2-2\left(a-b\right)\right]\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left(a-b\right)^2+4\left(a-b\right)\)
\(=4\)
b: \(\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{128}-1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(=2^{512}-1+1=2^{512}\)
c: \(24\left(5^2+1\right)\left(5^4+1\right)\cdot...\cdot\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}\)
=-1
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 .
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\frac{100.\left(100+1\right)}{2}=5050\)
b) \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=...=\left(2^{64}-1\right)\left(2^{64}+1\right)+1^2=2^{128}-1^2+1^2=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
\(=2c^2\)
a/Có A=100^2+99^2+98^2+...+1^2 -2(99^2+97^2+..+1)
= Sigma(100)(x=1)(x^2) -2((1^2+2^2+3^2+..+99^2)-(2^2+4^2+...+98^2)
=Sigma(100)(x=1)(x^2)-2.Sigma(99)(x=1)(x^2)+4sigma(49)(x=1)(x^2)
=5050
b/bạn lấy 3=2^2-1 rồi dùng hiệu 2 bình nhé
c/tách ra được thôi