Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\\ A=100+99+99+98+...+2+1\\ A=\left(100+1\right)\left(100-1+1\right):2=5050\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^1-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
\(C=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\\ C=2c^2\)
a: \(A=\left(100-99\right)\left(100+99\right)+\left(98+97\right)\left(98-97\right)+....+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+...+2+1\)
=5050
b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\cdot\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1=2^{128}\)
a. \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=199+195+...+3\)
\(=\dfrac{\left(199+3\right)\left(\dfrac{199-3}{4}+1\right)}{2}=5050\)
b. \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=2^{128}-1+1=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-2b^2-4ab\)
\(=2c^2\)
a: A=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)
=100+99+98+...+2+1
=5050
b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)+1
\(=2^{64}-1+1=2^{64}\)
a, A = 1002 - 992 + 982 - 972 +...+ 22 - 12
A = (1002 - 992) + (982 - 972) +...+ (22 - 1)2
A = (100 - 99)(100+99) + (98-97)(98+97)+..+(2-1)(2+1)
A = 1.199 + 1.195 + 1.191 +...+1.3
A = 3 + ...+191+ 195 + 199
Dãy số trên là dãy số cách đều với khoảng cách là: 199 -195=4
Dãy số trên có số hạng là: (199 - 3): 4 + 1 = 50 (số )
A = (199 +3) \(\times\) 50 : 2 = 5050
a) \(=\left(127+73\right)^2=200^2=40000\)
b) \(=18^8-\left(18^8-1\right)=1\)
c) \(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+...+2+1=5050\)
d) biến đổi thành \(20^2-19^2+18^2-17^2+..+2^2-1^2\)
rồi giải ra như trên
A = 1002 - 992 + 982 - 972 + ...... + 22 - 12
= ( 100 - 99 ) ( 100 + 99 ) + ( 98 - 97 ) ( 98 + 97 ) + ......... + ( 2 - 1 ) ( 2 + 1 )
= 1 + 2 + 3 + ......... + 99 + 100
= ( 100 + 1 ) . 100 : 2 = 5050
B = 3 ( 22 + 1 ) ( 24 + 1 ) ... ( 264 + 1 ) + 12
= ( 22 - 1 ) ( 22 + 1 ) ( 24 + 1 ) ... ( 264 + 1 ) + 1
= ( 24 - 1 ) ( 24 + 1 ) ... ( 264 + 1 ) + 1
= ( 28 - 1 ) ( 28 + 1 ) ... ( 264 + 1 ) + 1
= ( 216 - 1 ) ( 216 + 1 ) ... ( 264 + 1 ) + 1
= ( 232 - 1 ) ( 232 + 1 ) ( 264 + 1 ) + 1
= ( 264 - 1 ) ( 264 + 1 ) + 1
= 2128 - 1 + 1
= 2128
C = ( a + b + c )2 + ( a + b - c )2 - 2 ( a + b )2
= a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + ab - ac + ab + b2 - bc - ac - bc + c2 - 2 ( a2 + 2ab + b2 )
= a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + ab - ac + ab + b2 - bc - ac - bc + c2 - 2a2 - 4ab - 2b2
= 2c2
Bài làm:
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{\left(100+1\right)\times100}{2}=5050\)
b) \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
...
\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(B=2^{128}-1+1=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(C=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2\left(ab-bc-ca\right)-2a^2-4ab-2b^2\)
\(=2c^2+2\left(ab+bc+ca+ab-bc-ca-2ab\right)\)
\(=2c^2+2.0=2c^2\)