Đặt \(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

\(A=1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+...+\frac{1+2+3+...+16}{16}\)

\(A=1+\frac{2\left(2+1\right):2}{2}+\frac{3\left(3+1\right):2}{3}+\frac{4\left(4+1\right):2}{4}+...+\frac{16\left(16+1\right):2}{16}\)

\(A=1+\frac{2+1}{2}+\frac{3+1}{2}+\frac{4+1}{2}+...+\frac{16+1}{2}\)

\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(A=\frac{2+3+4+5+...+17}{2}\)

\(A=\frac{152}{2}\)

\(A=76\)

A=1+\(\frac{1+2}{2}\)+\(\frac{1+2+3}{3}\)+...+\(\frac{1+2+3+...+16}{16}\)

A=\(\frac{2}{2}\)+\(\frac{3}{2}\)+\(\frac{4}{2}\)+...+\(\frac{17}{2}\)

A=\(\frac{2+3+4+...+17}{2}\)

A=76(đề thi HSG huyện tui có tui làm zậy mà cũng có điểm tuyệt đối)

\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{16}.\left(1+2+3+....+16\right)\)

\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+\frac{1}{4}\cdot\frac{4.5}{2}+.....+\frac{1}{16}\cdot\frac{16.17}{2}\)

\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{17}{2}\)

\(A=\frac{\left(2+3+4+.....+17\right)}{2}=\frac{\left(2+17\right).16}{2}=\frac{152}{2}=76\)

A = 1 + 1/2.(1+2) + 1/3.(1+2+3) + 1/4.(1+2+3+4) + ...+ 1/16.(1+2+....+16)

A = 1 + 1/2.3 + 1/3.6 + 1/4.10 + ...+ 1/16.136

A = 1 + 3/2 + 4/2 + 5/2 + ....+ 17/2

A = 1 + (3+4+5+...+17)/2

A = 1 + 150/2

A = 1 + 75

A = 76

A = 1 + 1/2.(1+2) + 1/3.(1+2+3) + 1/4.(1+2+3+4) + ...+ 1/16.(1+2+....+16)

A = 1 + 1/2.3 + 1/3.6 + 1/4.10 + ...+ 1/16.136

A = 1 + 3/2 + 4/2 + 5/2 + ....+ 17/2

A = 1 + (3+4+5+...+17)/2

A = 1 + 150/2

A = 1 + 75

A = 76

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{16}\left(1+2+3+....+16\right)\)

\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+....+\frac{1}{16}\cdot\frac{16.17}{2}\)

\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{17}{2}\)

\(A=\frac{\left(2+3+4+....+17\right)}{2}=\frac{\left(2+17\right).\left(17-2+1\right):2}{2}=\frac{152}{2}=76\)

Ta có: \(A = 1+{1+2\over 2} + {1+2+3\over 3} +...+{1+2+...+ 16\over 16}\)

Xét: \(S_n = 1+2+3+...+n =\frac{n(n+1)}{n} (n \in N^*)\)

=> \({S_n\over n} = {(n+1)\over 2}\)

Thay vào biểu thức A ta có: 

\(A=1 + {3\over 2} + {4\over 2} + ... + {17\over 2}\)

\(A={(2+3+4+...+17)\over 2}\)

\(A={(17+2)[(17-2+1):2]\over 2} = {152\over2}=76\)

x. (x^2)^3 = x^5 
x^7 ≠ x^5 
Nếu, 
x^7 - x^5 = 0 
mủ lẻ nên phương trình có 3 nghiệm 
Đáp số: 
x = -1 
hoặc 
x = 0 
hoặc 
x = 1 

a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)

\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)

\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)

\(=\frac{7}{12}\)

b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)

\(=\frac{1}{5}\)

\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+...+16\right)\)

\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{16}.16.17:2=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}=\frac{2+3+4+...+17}{2}=\frac{152}{2}=76\)

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(=\frac{\frac{17.18}{2}-1}{2}=76\)