A = 100² - 99² + 98² - 97² + . . . + 2² - 1²

= (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + . . . (2 - 1)(2 + 1)

= 199 + 195 + . . . + 3

= 5050

B = 3(2² + 1)(2⁴ + 1) . . . (2⁶⁴ + 1) + 1

= (2² - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1)(2⁶⁴ + 1) + 1

= (2⁴ - 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2⁸ - 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2¹⁶ - 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2³² - 1)(2³² + 1)(2⁶⁴ + 1) + 1

= (2⁶⁴ - 1)(2⁶⁴ + 1) + 1

= 2¹²⁸ - 1 + 1

= 2¹²⁸

Câu C mk chép nhầm đề đó

a) ( x² - 2x + 2 )( x² - 2 )( x² + 2x + 2 )( x² + 2 )

= [ ( x² + 2 )² - 4x² ] ( x⁴ - 4 )

= ( x⁴ + 4 ) ( x⁴ - 4 )

= x⁸ - 16

b) ( a + b + c )² + ( a + b - c )² + ( 2a -b )²

= 2 ( a² + b² + c² ) + 2 ( ab + bc + ac ) + 2 ( ab - bc - ac ) + ( 4a² - 4ab + b² )

= 2 ( a² + b² + c² ) + 4ab - 4ab + 4a² + b²

= 6a² + 3b² + 2c²

c) 100² - 99² + 98² - 97² + ..... + 2² - 1²

= ( 100 - 99 )( 100 + 99 ) + ( 98 - 97 )( 98 + 97 ) + ..... + ( 2 - 1 )( 2 + 1 )

= 199 + 197 + 195 + ..... + 5 + 3

= \(\frac{\left(199+3\right)\left(\left(199-3\right)\frac{1}{2}+1\right)}{2}\)

= 9999

d) 3 ( 2² + 1 )( 2⁴ +1 )......( 2⁶⁴ + 1 ) + 1

= ( 2² -1 )( 2² + 1 )(2⁴ + 1 ).....( 2⁶⁴ + 1 ) + 1

= ( 2⁴ -1 )( 2⁴ + 1 )( 2⁸ + 1 )......( 2⁶⁴ + 1 ) +1

= ( 2⁸ -1 )( 2⁸ + 1).....( 2⁶⁴ + 1) +1

............

= ( 2⁶⁴ - 1)( 2⁶⁴ +1 ) + 1

= 2¹²⁸

a,b,c,f tìm cách áp dụng HĐT vào nhé! động não tí xem :)

d) Sửa đề :\(100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)

\(=199+195+...+3\)

Khi đó tổng sẽ là:

\(\dfrac{\left(199+3\right)\left[\dfrac{\left(199-3\right)}{4}+1\right]}{2}=5050.\)

e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1\)

\(=2^{128}.\)

lười quá, ko mún tính^^

nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)

\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=\left(2a+2b\right)^2-2\left(a+b+c\right)\left(a+b-c\right)-2\left(a+b\right)^2\)

\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)

\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)

g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)

\(=\left(2a+2b\right)^2-2\left(a+b+c+d\right)\left(a+b-c-d\right)+\left(2a-2b\right)^2-2\left(a+c-b-d\right)\left(a+d-b-c\right)\)

\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)

\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)

\(=2\left(\left(2a\right)^2-2\left(a+b\right)\left(a-b\right)\right)+2\left(\left(2c\right)^2-2\left(c+d\right)\left(c-d\right)\right)\)

\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)

\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)

bn đăng nhiều quá nên mk làm câu nào hay câu đó nha

mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .

a/ A = 100² - 99² + 98² -...+2² - 1²

= (100² - 99²) + (98² - 97²) +...+ (2² - 1²)

= 199 + 195 + 191 + ... + 1

= (\(\frac{199-1}{4}+1\))(\(\frac{199+1}{2}\)) = 5050

b/ Y chang câu a luôn nha

c/ \(C=\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}\)

\(=\frac{560.1000}{200^2}=14\)

d/ 2(a + 4)(a + 5) - (a + 5)² - (a² - 9a - 17) 

= 17a + 32 = 17(a + 1) + 15 = 1700 + 15 = 1715

Bài 1:

a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)

b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(18^8-\left(18^8-1\right)=1\)

\(c,100^2-99^2+98^2-97^2+...+2^2-1\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)

áp dụng công thức Gauss ta đc đáp án là:10100

d, mk khỏi ghi đề dài dòng:

\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:

\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)

\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)

\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)

1c,

\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)