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a. A = 1002 - 992+ 982 - 972 + ... + 22 - 12
A = ( 1002 - 992 ) + ( 982 - 972 ) + ... + ( 22 - 12 )
A = ( 100 - 99 )(100 + 99 ) + (98 - 97 )(98 + 97) + ... + (2-1)(2+1)
A = 199 + 195 + .... + 3
Tổng A có ss hạng là:
( 199 - 3 ) : 4 + 1 = 50 ( số )
Tổng A bằng:
( 199 + 3 ) x 50 : 2 = 5050
c. C = (a + b + c)2 + (a + b - c)2 - 2(a + b)2
C = a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + b2 + c2 + 2ab - 2bc - 2ac - 2(a2 + 2ab + b2)
C = 2a2 + 2b2 + 2c2 + 4ab - 2a2 -4ab - 2b2
C = 2c2
b. B = 3(22 + 1) (24 + 1) ... (264 + 1) + 12
B = (22 - 1)(22 + 1)(24 + 1) ... (264 + 1) + 12
B = ( 24 - 1)(24 + 1)... (264 + 1) + 12
B = (28 - 1)... (264 + 1) + 12
B = (28 - 1)(28+1)... (264 + 1) + 12
B = (216-1)(216+1)... (264 + 1) + 12
B = (232 - 1)(232+1)... (264 + 1) + 12
B = (264 - 1)(264 +1)+1
B = 2128 - 1 + 1
B = 2128
\(A=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\\ A=100+99+99+98+...+2+1\\ A=\left(100+1\right)\left(100-1+1\right):2=5050\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^1-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
\(C=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\\ C=2c^2\)
Cũa mị:>>>
Tham khảo ạ !!!
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
a) \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+...+2+1\)
\(=\frac{100\left(100+1\right)}{2}=5050\)
Cái này là chứng minh VT=VP đk?
a)\(a^3+b^3=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)-\left(3a^2b+3ab^2\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)\)
b)Mk ko bt làm !
Ta có : A = 1002 - 992+ 982 - 972 + ... + 22 - 12
=> A = (1002 - 992) + (982 - 972) + ...... + (22 - 12)
=> A = (100 + 99)(100 - 99) + (98 + 97)(98 - 97) + ..... + (2 + 1)(2 - 1)
=> A = 199 + 195 + 191 + .... + 3
=> A = SSH.(199 + 3) : 2
Trong đó SSH = (199 - 3) : 4 + 1 = 50 số
Thay vào A ta có : A = 50(199 + 3) : 2 = 5050.
b. \(3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4-1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1\)
\(=2^{128}\)
Hằng đẳng thức mk dùng là \(\left(a+b\right)\left(a-b\right)=a^2-b^2\)nhé
a, A = (100-99).(100+99)+(98-87).(98+97)+....+(2-1).(2+1)
= 199.1+195.1+...+3.1 = 199+195+....+3 = (199+3).[ (199-3):4+1 ] :2 = 202 . 50 : 2 = 5050
k mk nhja
a, \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\frac{\left(1+100\right).100}{2}=5050\)
a) \(\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=a^3+b^3\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ac-3bc-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).
a) Ta có: \(\left(a+b\right)^3=a^3+b^3+3a^2b+3ab^2\)
\(=a^3+b^3+3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)(đpcm)
b) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\left(a+b+c\right)\right]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)(đpcm)