\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{198}{2}+1\right)+1}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{200\cdot\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right)}\)

\(=\frac{1}{200}\)

Câu a

\(S=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{2019-2017}{2017x2019}.\)

\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}=1-\frac{1}{2019}=\frac{2018}{2019}\)

Câu b

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^6}+\frac{1}{3^7}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^5}+\frac{1}{3^6}\)

\(2A=3A-A=1-\frac{1}{3^7}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^7}\)

\(X=\frac{1}{32}\)

x = 1/32

\(\frac{5}{8}=0,625;\frac{7}{10}=0,7\)

vì \(0,625< 0,7\)NÊN \(\frac{5}{8}< \frac{7}{10}\)

VẬY \(\frac{5}{8}< \frac{7}{10}\)

TK MN NHÉ

Quy đồng, ta có:

\(\frac{5}{8}=\frac{25}{40}\)

\(\frac{7}{10}=\frac{28}{40}\)

Vì \(25< 28\)nên \(\frac{25}{40}< \frac{28}{40}\)

 \(\Rightarrow\frac{5}{8}< \frac{7}{10}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

\(A=2A-A=1-\frac{1}{2^{10}}\Rightarrow A+\frac{1}{2^{10}}=1-\frac{1}{2^{10}}+\frac{1}{2^{10}}=1\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\)

\(A+\frac{1}{2^{10}}=1\)

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

\(\frac{2}{3}.x=3-\frac{3}{4}\)

\(\frac{2}{3}.x=\frac{9}{4}\)

\(x=\frac{9}{4}:\frac{2}{3}\)

\(x=\frac{27}{8}\)

\(\frac{2}{3}\cdot x+\frac{3}{4}=3\)

\(\frac{2}{3}\cdot x=3-\frac{3}{4}\)

\(\frac{2}{3}\cdot x=\frac{9}{4}\)

         \(x=\frac{9}{4}:\frac{2}{3}\)

         \(x=\frac{27}{8}\)

giúp mik ik ạk

\(a+b=132\)\(\left(1\right)\)

\(a-b=4\) \(\left(2\right)\)

lấy \(\left(1\right)-\left(2\right)\)ta có 

\(a+b-a+b=132-4\)

<=> \(2b=128\)

<=> \(b=64\)

=> \(a=4+b=4+64=68\)

\(a+b=132\)

\(a-b=4\)

\(\Rightarrow a=\left(132+4\right)\text{ : }2\)

\(\Rightarrow a=136\text{ : }2\)

\(\Rightarrow a=68\)

\(\Rightarrow b=68-4\)

\(\Rightarrow b=64\)

\(\text{Vậy : }a=68\)

            \(b=64\)

\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{16}=\left(\frac{21}{8}+\frac{4}{8}\right)\cdot\frac{16}{5}\)

                                  \(=\frac{25}{8}\cdot\frac{16}{5}\)

                                      \(=10\)

\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{16}\)

\(=\frac{25}{8}:\frac{5}{16}\)

\(=10\)

#Học Tốt