ta có 50^40>50^39

        50^39>50^38=)1/50^39<1/50^38

=)50^40+1/50^39>50^39+1/50^38

=)50^40+1/50^39+1>50^39+1/50^38+1

=)A>B

Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)

Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)

\(\dfrac{1}{32}>\dfrac{1}{40}\)

\(\dfrac{1}{33}>\dfrac{1}{40}\)

\(\dfrac{1}{34}>\dfrac{1}{40}\)

\(\dfrac{1}{35}>\dfrac{1}{40}\)

\(\dfrac{1}{36}>\dfrac{1}{40}\)

\(\dfrac{1}{37}>\dfrac{1}{40}\)

\(\dfrac{1}{38}>\dfrac{1}{40}\)

\(\dfrac{1}{39}>\dfrac{1}{40}\)

\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)

Vậy \(S>\dfrac{1}{4}\)

a) \(9^3\cdot3^2\)

\(=\left(3^2\right)^3\cdot3^2\)

\(=3^6\cdot3^2\)

\(=3^8\)

b) \(x^7\cdot x:x^4\)

\(=x^8:x^4\)

\(=x^4\)

c) \(7\cdot3^9+3^{10}+51\cdot3^8\)

\(=3^8\cdot\left(7\cdot3+3^2+51\right)\)

\(=3^8\cdot81\)

\(=3^8\cdot3^4\)

\(=3^{12}\)

d) \(5^{15}\cdot125^3\cdot625\)

\(=5^{15}\cdot5^9\cdot5^4\)

\(=5^{28}\)

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

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

=>\(\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{20}>\frac{10}{20}=\frac12\) (1)

Ta có: \(\frac{1}{21}>\frac{1}{30};\frac{1}{22}>\frac{1}{30};\ldots;\frac{1}{30}=\frac{1}{30}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}\)

=>\(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}>\frac{10}{30}=\frac13\) (2)

Ta có: \(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};\ldots;\frac{1}{40}=\frac{1}{40}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}\)

=>\(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}>\frac{10}{40}=\frac14\) (3)

Từ (1),(2),(3) suy ra \(\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}\right)>\frac12+\frac13+\frac14\)

=>A>13/12

=>A>1

=>A>2012/2013

\(\frac{195}{1890}>\frac{39}{379}\)

\(\frac{193}{310}>\frac{173}{350}\)

\(k-mk-nha\)

\(A=\frac{7^{38}+10}{7^{38}-3}\)như thế này có đúng không bạn???

Ta có: \(A=\frac{7^{38}+10}{7^{38}-3}=\frac{7^{38}-3+13}{7^{38}-3}=1+\frac{13}{7^{38}-3}\)

Lại có: \(B=\frac{7^{39}+18}{7^{39}+5}=\frac{7^{39}+5+13}{7^{39}+5}=1+\frac{13}{7^{39}+5}\)

Vì \(\frac{13}{7^{38}-3}>\frac{13}{7^{39}+5}\) nên \(1+\frac{13}{7^{38}-3}>1+\frac{13}{7^{39}+5}\)

\(\Rightarrow A>B\)

Vậy \(A>B\).

cảm ơn bạn nhưng mik xong rùi:>