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\(A=\left(\frac{20}{5}+\frac{27}{9}\right)\times\frac{21}{10}=\left(4+3\right)\times\frac{21}{10}=7\times\frac{21}{10}=\frac{147}{10}\)
\(B=\left(\frac{13}{6}-\frac{3}{8}\right)\times\frac{11}{22}\)
\(B=\left(\frac{52}{24}-\frac{9}{24}\right)\times\frac{11}{22}\)
\(B=\frac{43}{24}\times\frac{1}{2}=\frac{43}{48}\)
Dễ thấy \(A=\frac{147}{10}>1\)
Mà \(B=\frac{43}{48}< 1\)
=> tự so sánh
\(\dfrac{21}{36}-\left(-\dfrac{11}{30}\right)=\dfrac{7}{12}+\dfrac{11}{30}=\dfrac{7.5+11.2}{60}=\dfrac{57}{60}=\dfrac{19}{20}\\ ----\\\dfrac{-4}{8}+\left(-\dfrac{3}{10}\right)=\dfrac{-1}{2}-\dfrac{3}{10}=\dfrac{-1.5-3}{10}=\dfrac{-8}{10}=-\dfrac{4}{5}\\ ----\\ \dfrac{7}{12}-\left(-\dfrac{9}{20}\right)=\dfrac{7}{12}+\dfrac{9}{20}=\dfrac{7.5+9.3}{60}=\dfrac{62}{60}=\dfrac{31}{30}\\ ---\\ \dfrac{-2}{5}+\left(-\dfrac{11}{30}\right)=-\dfrac{2}{5}-\dfrac{11}{30}=\dfrac{-2.6-11}{30}=-\dfrac{29}{30}\)
Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
A = \(\dfrac{10^{20}+3}{10^{21^{ }}+3}\)
B = \(\dfrac{10^{21}+4}{10^{22}+4}\) < 1
\(\Rightarrow\) B < \(\dfrac{10^{21}+4+6}{10^{22}+4+6}\)
\(\Rightarrow\) B < \(\dfrac{10^{21}+10}{10^{22}+10}\)
\(\Rightarrow\) B < \(\dfrac{10\left(10^{20}+1\right)}{10\left(10^{21}+1\right)}\)
\(\Rightarrow\) B < \(\dfrac{10^{20}+1}{10^{21}+1}\) < \(\dfrac{10^{21}+1+2}{10^{22}+1+2}\)
\(\Rightarrow\) B < \(\dfrac{10^{21}+3}{10^{22}+3}\)
\(\Rightarrow\) B < A