Ta có:

M=\(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\)

M=\(\frac{1.3....99}{2.4....100}\)

Lại có:

N=\(\frac{2}{3}.\frac{4}{5}....\frac{100}{101}\)

N=\(\frac{2.4....100}{3.5....101}\)

\(\Rightarrow\)M.N=\(\frac{1.2.3......99.100}{2.3.4......100.101}\)

\(\Rightarrow\)M.N=\(\frac{1}{101}\)

Câu trả lời hay nhất:  k² > k² - 1 = (k-1)(k+1) 
⇒ 1/k² < 1/[(k-1).(k+1)] = [1/(k-1) - 1/(k+1)]/2 (*) 

Áp dụng (*), ta có: 
1/2² + 1/3² + 1/4² + ... + 1/n² 
< 1/2² + 1/(2.4) + 1/(3.5) + ... + 1/[(n-1).(n+1)] 
= 1/2² + [1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-1) - 1/(n+1)]/2 
= 1/2² + [1/2 + 1/3 - 1/n - 1/(n+1)]/2 
= 2/3 - [1/n + 1/(n+1)]/2 < 2/3

1.

Ta có:

1/2 < 2/3

3/4 < 4/5

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

99/100 < 100/101

=> 1/2*3/4*5/6*...*99/100 < 2/3*4/5*6/7*...*100/101

=> A < B

2.

\(A\cdot B=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)

\(A\cdot B=\frac{\left[1\cdot3\cdot5\cdot7\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot9\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)

3.

Vì A < B => A.A < A.B => A² < 1/101 < 1/100

Mà A² < 1/100 <=> A² < \(\frac{1}{10}^2\)=> A < 1/10

Bài 1 : tìm N thuộc N , biết :

a) 1<2^n < 128

2^0 < 2^n < 2^7

0 < n < 7

Vì n là số tự nhiên nên n = 1; 2; 3; 4; 5; 6

Vậy n ∈ {1; 2; 3; 4; 5; 6}

b) 9 < 3^n < 729

2^2 < 3^n < 3^6

2 < n < 6

Vì n là số tự nhiên nên n = 3; 4; 5

Vậy n ∈ {3; 4; 5}

c) 1 <=3^2n <= 27 ^ 2

3^0 ≤ 3^2n ≤ 3^6

0 ≤ 2n ≤ 6

0 ≤ n ≤ 3

Vì n là số tự nhiên nên n ∈ {0; 1 ; 2; 3}

Minh lam cau A) thoi duoc hong

chắc là bạn ghi sai đề rồi

bạn gửi = máy tính hay điện thoại mà đề bài sai tùm lum zậy?

a: \(P=5+5^2+5^3+5^4+\cdots+5^{102}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{101}+5^{102}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{101}\left(1+5\right)\)

\(=6\left(5+5^3+\cdots+5^{101}\right)\) ⋮6

b:Sửa đề: \(A=1+4+4^2+4^3+\cdots+4^{99}\)

\(=\left(1+4\right)+\left(4^2+4^3\right)+\cdots+\left(4^{98}+4^{99}\right)\)

\(=\left(1+4\right)+4^2\left(1+4\right)+\cdots+4^{98}\left(1+4\right)\)

\(=5\left(1+4^2+\cdots+4^{98}\right)\) ⋮5

c: \(B=1+2+2^2+\cdots+2^{98}\)

\(=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+\cdots+\left(2^{96}+2^{97}+2^{98}\right)\)

\(=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+\cdots+2^{96}\left(1+2+2^2\right)\)

\(=7\left(1+2^3+\cdots+2^{96}\right)\) ⋮7

d:Sửa đề: \(C=1+3+3^2+3^3+\cdots+3^{103}\)

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\cdots+\left(3^{100}+3^{101}+3^{102}+3^{103}\right)\)

\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+\cdots+3^{100}\left(1+3+3^2+3^3\right)\)

\(=40\left(1+3^4+\cdots+3^{100}\right)\) ⋮40