nhiều thế, đăng ít một thôi bạn

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

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

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

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

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

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

A = -1² + 2² - 3² + 4² - ... - 99² + 100²

A = 100² - 99² + ... + 4² - 3² + 2² - 1²

A = (100 + 99).(100 - 99) + ... + (4 + 3).(4 - 3) + (2 + 1).(2 - 1)

A = 100 + 99 + ... + 4 + 3 + 2 + 1

\(A=\frac{\left(1+100\right).100}{2}=101.50=5050\)

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)

2B = (3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)...(3³² + 1)

2B = (3² - 1)(3² + 1)(3⁴ + 1)...(3³² + 1)

2B = (3⁴ - 1)(3⁴ + 1)...(3³² + 1)

2B = 3⁶⁴ - 1

\(B=\frac{3^{64}-1}{2}\)

A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n⁴ + 4 = (n²+2)² - 4n² = (n²-2n+2)(n²+2n+2) = [(n-1)²+1][(x+1)²+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

b) -1² + 2² - 3² + 4² - ... - 99² + 100²

= (2² - 1²) + (4² - 3²) + ... + (100² - 99²)

= (2 + 1)(2 - 1) + (4 + 3)(4 - 3) + ... + (100 + 99)(100 - 99)

= (1 + 2) + (3 + 4) + ... + (99 + 100)

= 5050

a) (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)

= [(3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)] : 2

= [(3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)] : 2

= [(3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)] : 2

Và cứ như thế ta được kết quả là (3³² - 1) : 2 = 926510094425920

b: Để N là số nguyên dương thì \(\sqrt{x}-3>0\)

\(\Leftrightarrow x>9\)

mà x là số nguyên

nên \(\left\{{}\begin{matrix}x\in Z\\x>9\end{matrix}\right.\)

a)\(T=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)

ta có \(2+1=2^2-1\)

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

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

\(T=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(T=2^{32}-1\)

bạn ơi nơi chổ mấy cái  \(\left(2^2-1\right)\left(2^2+1\right)\)là nhân đa thức lại nha

b)

\(U=100^2-99^2+98^2-97^2+...+4^2-3^2+2^2-1^2\)

\(U=-1^2+2^2-3^2+4^2-...-97^2+98^2-99^2+100^2\)

\(U=2^2-1^2+4^2-3^2+...+98^2-97^2+100^2-99^2\)

\(U=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)(dùng hằng đẳng thức sô 3 nha)

\(U=3+7+...+199\)

\(U=1+2+3+\text{4+...+99+100}\)

số số hạng của U là :\(\left(100-1\right):1+1=100\) (số hạng)

tổng số số hạng của U là : \(\frac{\left(100+1\right).100}{2}=5050\)

à bạn coi lại cái đề nha đoạn sau hình như thiếu 2^2 thì phải

a) \(\left(n-1\right)^2-n\left(n-2\right)=3\left(n-1\right)\)

\(\Rightarrow n^2-2n+1-n^2+2n=3n-3\)

\(\Rightarrow3n-3=1\)

\(\Rightarrow3n=4\)

\(\Rightarrow n=\dfrac{4}{3}\)

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