A= 101-100+99-98+.......3-2+1

A= (101-100) + (99-98) + ... + (5-4) + (3-2) +1 
A= 1 + 1 + ... + 1 + 1 + 1 
A= 1 x 51 
A= 51

A = 101 - 100 + 99 - 98 + ....... 3 - 2 + 1

A = ( 101 - 100 ) + ( 99 - 98 ) + ... + ( 5 - 4 )  + ( 3 - 2 ) +1 
A = 1 + 1 + ... + 1 + 1 + 1 
A = 1 x  51 
A = 51

\(\frac{\left(5^4-5^3\right)^3}{125^4}\)

\(=\frac{500^3}{125^4}\)

\(=\frac{\left(125.4\right)^3}{125^4}\)

\(=\frac{4^3}{125}=\frac{4^3}{5^3}=\left(\frac{4}{5}\right)^3\)

a) \(\frac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\frac{2^{23}}{2^{24}}=\frac{1}{2}\)

Các phần kia tương tự, à bạn đăng 1 2 câu hỏi 1 lần thôi, đăng nhiều quá ko ai trả lời đâu

@-@

Thế à ! Vậy bạn hãy nhấp vào https://h.vn/hoi-dap/question/646555.html?pos=1792187 mà xem

\(\frac{3}{15}\cdot G=\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{68\cdot71}\)

\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)

\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{71}\)

\(G=\frac{60}{781}\cdot\frac{15}{3}\)

\(G=\frac{300}{781}\)

ta có :\(\frac{3}{15}G=\left(\frac{15}{11.14}+\frac{15}{14.17}+...+\frac{15}{68.71}\right)\)

\(\frac{3}{15}G=\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{68.71}\)

\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)

\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{71}=\frac{71}{781}-\frac{11}{781}=\frac{60}{781}\)

\(=>G=\frac{60}{781}:\frac{3}{15}=\frac{900}{2343}\)

vậy G =900/2343