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Ta có: \(\frac{A}{10^{10}}=\frac{10^{20}-6}{10^{20}-6\cdot10^{10}}=\frac{10^{20}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}\)
\(\frac{B}{10^{10}}=\frac{10^{21}-6}{10^{21}-6\cdot10^{10}}=\frac{10^{21}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
Ta có: \(10^{20}<10^{21}\)
=>\(10^{20}-6\cdot10^{10}<10^{21}-6\cdot10^{10}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}+1>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}+1\)
=>\(\frac{A}{10^{10}}>\frac{B}{10^{10}}\)
=>A>B


c ) 5 / x - 4 / 3 = 1 / 6
5 / x = 4 / 3
5 . 3 = x . 4
15 = x . 4
x = 15 . 4
x = 60
5 / x = 1 / 6
5 . 6 = x . 1
30 = x . 1
x = 30 . 1
x = 30

Ta có: \(10A=\frac{10^{21}-60}{10^{21}-6}=\frac{10^{21}-6-54}{10^{21}-6}=1-\frac{54}{10^{21}-6}\)
\(10B=\frac{10^{22}-60}{10^{22}-6}=\frac{10^{22}-6-54}{10^{22}-6}=1-\frac{54}{10^{22}-6}\)
Ta có: \(10^{21}-6<10^{22}-6\)
=>\(\frac{54}{10^{21}-6}>\frac{54}{10^{22}-6}\)
=>\(-\frac{54}{10^{21}-6}<-\frac{54}{10^{22}-6}\)
=>\(-\frac{54}{10^{21}-6}+1<-\frac{54}{10^{22}-6}+1\)
=>10A<10B
=>A<B

\(a,3x+17=-5\\ \Rightarrow3x=-22\\ \Rightarrow x=-\dfrac{22}{3}\\ b,\dfrac{-5}{8}=\dfrac{x}{16}\\ \Rightarrow x=\dfrac{-5}{8}.16\\ \Rightarrow x=-10\\ c,\dfrac{4}{9}+\dfrac{7}{3}:x=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{3}:x=-\dfrac{11}{45}\\ \Rightarrow x=-\dfrac{105}{11}\)
\(a)3.x+17=-5\)
\(3.x\) \(=\left(-5\right)-170=-22\)
\(x\) \(=\left(-22\right)\div3\)
\(x\) \(=\dfrac{-22}{3}\)
\(b)\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{\left(-5\right)\times16}{8}=\dfrac{-80}{8}=-10\)
\(c)\dfrac{4}{9}+\dfrac{7}{3}:x=\dfrac{1}{5}\)
\(\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}\)
\(\dfrac{7}{3}:x=\dfrac{1}{5}+\left(\dfrac{-4}{9}\right)=\dfrac{9}{45}+\left(\dfrac{-20}{45}\right)=\dfrac{-11}{45}\)
\(x=\dfrac{7}{3}:\left(\dfrac{-11}{45}\right)=\dfrac{7}{3}\times\left(\dfrac{-45}{11}\right)\)
\(x=\dfrac{-315}{33}=\dfrac{-105}{11}\)