a: \(\left(\dfrac{1}{5}\right)^{-2}=25\)

b: \(4^{\dfrac{3}{2}}=8\)

c: \(\left(\dfrac{1}{8}\right)^{-\dfrac{2}{3}}=\left(\dfrac{1}{2}\right)^{3\cdot\dfrac{-2}{3}}=\left(\dfrac{1}{2}\right)^{-2}=4\)

d: \(\left(\dfrac{1}{16}\right)^{-0.75}=\left(\dfrac{1}{2}\right)^{4\cdot\left(-0.75\right)}=\left(\dfrac{1}{2}\right)^{-3}=8\)

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

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

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

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

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: 

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

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

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

\(=1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2}+...+8+\frac{17}{2}\)

\(=\left(1+2+...+8\right)+\left(\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\right)=36+\frac{80}{2}=36+40=76\)

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

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

\(B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)

\(B=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(B=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)

\(B=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)

Không thấy gì hết

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)

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

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

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

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

Ế, bạn Phạm Nguyễn Tất Đạt, bạn đâu có được điểm tuyệt đối

a)

\(\begin{array}{l}{\left( {1 + \frac{1}{2} - \frac{1}{4}} \right)^2}.\left( {2 + \frac{3}{7}} \right)\\ = {\left( {\frac{4}{4} + \frac{2}{4} - \frac{1}{4}} \right)^2}.\left( {\frac{{14}}{7} + \frac{3}{7}} \right)\\ = {\left( {\frac{5}{4}} \right)^2}.\frac{{17}}{7}\\ = \frac{{25}}{{16}}.\frac{{17}}{7}\\ = \frac{{425}}{{112}}\end{array}\)

b)

\(\begin{array}{l}4:{\left( {\frac{1}{2} - \frac{1}{3}} \right)^3}\\ = 4:{\left( {\frac{3}{6} - \frac{2}{6}} \right)^3}\\ = 4:{\left( {\frac{1}{6}} \right)^3}\\ = 4:\frac{1}{{216}}\\ = 4.216\\ = 864\end{array}\)

Ta có công thức : 1 + 2 + 3 + ... + n = \(\frac{n\left(n+1\right)}{2}\)

Do đó

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

\(P=1+\frac{2.3}{2.2}+\frac{3.4}{2.3}+\frac{4.5}{2.4}+...+\frac{16.17}{2.16}\)

\(P=1+\frac{1}{2}\left(3+4+5+...+17\right)\)

\(P=1+\frac{1}{2}.\frac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)

Xét thừa số tổng quát:

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

Thay vào bài toán:

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

\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}=\frac{2+3+...+17}{2}=76\)