c: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{49\cdot50}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{49}+\frac{1}{50}-2\left(\frac12+\frac14+\cdots+\frac{1}{50}\right)\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{50}-1-\frac12-\cdots-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\)

giúp em câu a b nx dc hem tại khó quá em chx học kiểu chấm than ở mẫu số

\(=\frac{47}{6}\)

Ta thấy :

1/1x2 = 1/1 - 1/2

1/2x3 = 1/2 - 1/3 

....

=>( 1/1x2 + 1/2x3 + 1/3x4 + 1/5x6 ) x 10 - x = ( 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ) x 10 - x

= ( 1/1 - 1/6 ) x 10 - x =0

 5/6 x 10 - x = 0 

 25/3 - x = 0

=> x = 25/3

( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0

= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0

= ( 1 - 1/6 ) x 10 - x = 0

= 5/6 x 10 - x =0

=   25/3 - x =0

               x = 25/3 - 0

                x  = 25/3

\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)

\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)

\(\frac{5}{6}\times10-x=0\)

\(\frac{25}{3}-x=0\)

x              =\(\frac{25}{3}-0=\frac{25}{3}\)

a. \(\frac{20^5.5^{10}}{100^5}\)

\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)

\(=\frac{20^5.25^5}{100^5}\)

\(=\frac{500^5}{100^5}\)

\(=\left(\frac{500}{100}\right)^5\)

\(=5^5=3125\)

b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)

\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)

\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)

\(=3^5.\frac{1}{0,3}\)

\(=810\)

c. \(\frac{6^3+3.6^2+3^3}{-13}\)

\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)

\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)

\(=\frac{3^3.13}{-13}\)

\(=\left(-3\right)^3\)

\(=-27\)

A=\(2.2^2+3.2^3+4.2^4+...+100.2^{100}\)

\(\Rightarrow2A=2.2^3+3.2^4+4.2^5+...+100.2^{101}\)

\(\Rightarrow A-2A=2.2^2+\left(3.2^3-2.2^3\right)+\left(4.2^4-3.2^4\right)+...+\left(100.2^{100}-99.2^{100}\right)-100.2^{101}\)

\(\Rightarrow-A=2^3+\left(2^3+2^4+...+2^{100}\right)-100.2^{101}\)

Đặt \(B=\left(2^3+2^4+...+2^{100}\right)\)

\(\Rightarrow2B=\left(2^4+2^5+...+2^{101}\right)\)

\(\Rightarrow2B-B=\left(2^4+2^5+...+2^{101}\right)-\left(2^3+2^4+...+2^{100}\right)\)

\(\Rightarrow B=2^{101}-2^3\)

\(\Rightarrow-A=2^3+2^{101}-2^3-100.2^{101}\)

\(\Rightarrow-A=2^{101}-100.2^{101}\)

\(\Rightarrow A=100.2^{101}-2^{101}=99.2^{101}\)