1 + 2 + 1 + 3 + 1 + 4 + .. + 1 + 100 

= 99 x 1 + ( 2 + 3 + 4 + ... + 100 )

= 5148          ( 1 )

1 + 3 + 5 +7 + ... +  301 

= \(\frac{\left[\left(301-1\right):2+1\right].\left(301+1\right)}{2}\)

= 22801          (2)

Từ ( 1) và (2) => 1+3+ ....+ 301 > 1+2+1+3+1+4 +...+ 1 + 100 

b) làm tương tự 

Câu đầu bé theo linh cảm thôi

Câu hai:Lớn vì phép đầu với phép hai ko có số 1,25 là bằng nhau nhưng lại có.

a < 1

Ta thấy: 1/6<1/5

               1/7<1/5

               1/8<1/5

               1/9<1/5

=>A=1/5+1/6+1/7+1/8+1/9<1/5+1/5+1/5+1/5+1/5

=>A<5.1/5

=>A<1

Vậy A<1

Ta có:

\(\frac{1}{5}=\frac{1}{5};\frac{1}{6}<\frac{1}{5};\frac{1}{7}<\frac{1}{5};\frac{1}{8}<\frac{1}{5};\frac{1}{9}<\frac{1}{5}\)

Nên\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}<\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1\)

Vậy A<1

a: Ta có: \(A=\frac59+\left(-\frac57\right)+\left(-\frac{20}{48}\right)+\frac{8}{12}+\left(-\frac{21}{48}\right)\)

\(=\frac59-\frac57-\frac{41}{48}+\frac{32}{48}\)

\(=\frac{35-45}{63}-\frac{9}{48}=\frac{-10}{63}-\frac{3}{16}=\frac{-160-189}{63\cdot16}=\frac{-349}{1008}\)

b: \(B=\left(-\frac59\right)+\frac{8}{15}+\left(-\frac{2}{11}\right)+\left(\frac{4}{-9}\right)+\frac{2}{45}\)

\(=\left(-\frac59-\frac49\right)+\frac{8}{15}+\frac{2}{45}-\frac{2}{11}\)

\(=-1-\frac{2}{11}+\frac{24}{45}+\frac{2}{45}=-\frac{13}{11}+\frac{26}{45}=\frac{-13\cdot45+26\cdot11}{11\cdot45}=\frac{-299}{495}\)

c: \(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\ldots;\frac{1}{20}=\frac{1}{20}\)

Do đó: \(\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}\)

=>S>10/20

=>S>1/2

\(A=\frac{7^{11}-7}{6}\)

\(B=\frac{5^{11}-5}{4}\)