Ta có: *)A.5=5(1/3.8+1/8.13+...+1/33.38)

                  =5/3.8+5/8.13+...+5/33.38

                  =1/3-1/8+1/8-1/13+...+1/33-1/38

                  =1/3-1/38

=> A=(1/3-1/38).1/5

*)7B=7/3.10+7/10.17+7/17.24+...+7/31.38

       =1/3-1/10+1/10-1/17+...+1/31-1/38

       =1/3-1/38

=>B=(1/3-1/38).1/7

Do đó a/b=(1/5)/(1/7)=7/5

k mk nha!

a) \(27^{64}:81^{20}=3^{192}:3^{80}=3^{112}\)

b) \(\left(\dfrac{1}{8}\right)^{20}:\left(\dfrac{1}{16}\right)^9=\left(\dfrac{1}{2}\right)^{60}:\left(\dfrac{1}{2}\right)^{36}=\left(\dfrac{1}{2}\right)^{24}\)

c) \(\dfrac{1}{3}:\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{3}-\dfrac{1}{6}=\dfrac{10}{6}-\dfrac{1}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)

i don'n know

Ta có: \(\frac{1}{151}>\frac{1}{300};\frac{1}{152}>\frac{1}{300};\ldots;\frac{1}{300}=\frac{1}{300}\)

Do đó: \(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+\cdots+\frac{1}{300}=\frac{150}{300}=\frac12\left(1\right)\)

Ta có: \(\frac{1}{301}>\frac{1}{450};\frac{1}{302}>\frac{1}{450};\ldots;\frac{1}{450}=\frac{1}{450}\)

Do đó; \(\frac{1}{301}+\frac{1}{302}+\cdots+\frac{1}{450}>\frac{1}{450}+\frac{1}{450}+\cdots+\frac{1}{450}=\frac{150}{450}=\frac13\) (2)

Từ (1),(2) suy ra \(\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)+\left(\frac{1}{301}+\frac{1}{302}+\cdots+\frac{1}{450}\right)\) >1/2+1/3

=>\(\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)+\left(\frac{1}{301}+\frac{1}{302}+\cdots+\frac{1}{450}\right)\) >5/6(ĐPCM)

\(a)\frac{x}{4}-\frac{3}{7}+\frac{2}{5}=\frac{31}{140}\)

\(\Leftrightarrow\frac{x}{4}-\frac{1}{35}=\frac{31}{140}\)

\(\Leftrightarrow\frac{x}{4}=\frac{1}{4}\Leftrightarrow x=1\)

\(b)\frac{5}{12}+\frac{5}{x}-\frac{1}{8}=\frac{1}{12}\)

\(\Leftrightarrow\frac{5}{x}+\frac{7}{24}=\frac{1}{12}\)

\(\Leftrightarrow\frac{5}{x}=-\frac{5}{24}\Leftrightarrow x=-24\)