c= (5^12+1)/(5^13+1) và D = (5^11+1)/(5^12+1) so sánh
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đặt A=\(\frac{5^{12}+1}{5^{13}+1}\);B=\(\frac{5^{11}+1}{5^{12}+1}\);C= \(\frac{5^{11}-1}{5^{12}-1}\)
ta có:nhân A,B,C với 5 ta đc:\(5A=\frac{5\left(5^{12}+1\right)}{5^{13}+1}=\frac{5^{13}+5}{5^{13}+1}=\frac{5^{13}+1+4}{5^{13}+1}=\frac{5^{13}+1}{5^{13}+1}+\frac{4}{5^{13}+1}=1+\frac{4}{5^{13}+1}\)
\(5B=\frac{5\left(5^{11}+1\right)}{5^{12}+1}=\frac{5^{12}+5}{5^{12}+1}=\frac{5^{12}+1+4}{5^{12}+1}=\frac{5^{12}+1}{5^{12}+1}+\frac{4}{5^{12}+1}=1+\frac{4}{5^{12}+1}\)
\(5C=\frac{5\left(5^{11}-1\right)}{5^{12}-1}=\frac{5^{12}-5}{5^{12}-1}=\frac{5^{12}-1-4}{5^{12}-1}=\frac{5^{12}-1}{5^{12}-1}-\frac{4}{5^{12}-1}=1-\frac{4}{5^{12}-1}\)
vì 513+1>512+1>512-1
=>\(\frac{4}{5^{12}-1}>\frac{4}{5^{12}+1}>\frac{4}{5^{13}+1}\)
\(\Rightarrow1+\frac{4}{5^{12}-1}>1+\frac{4}{5^{12}+1}>1+\frac{4}{5^{13}+1}\)
=>5C>5B>5A
=>C>B>A
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
công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
nên ta có : \(\frac{5^{12}+1}{5^{13}+1}< \frac{5^{12}+1+4}{5^{13}+1+4}\)\(=\frac{5^{12}+5}{5^{13}+5}=\frac{5.\left(5^{11}+1\right)}{5.\left(5^{12}+1\right)}=\frac{5^{11}+1}{5^{12}+1}\)
=> \(\frac{5^{12}+1}{5^{13}+1}< \frac{5^{11}+1}{5^{12}+1}\)
a) Đặt A = \(\frac{5^{12}+1}{5^{13}+1}\Rightarrow5A=\frac{5^{13}+5}{5^{13}+1}=1+\frac{4}{5^{13}+1}\)
Đặt \(B=\frac{5^{11}+1}{5^{12}+1}\Rightarrow5B=\frac{5^{12}+5}{5^{12}+1}=1+\frac{4}{5^{12}+1}\)
Vì \(\frac{4}{5^{13}+1}< \frac{4}{5^{12}+1}\Rightarrow1+\frac{4}{5^{13}+1}< 1+\frac{4}{5^{12}+1}\Rightarrow5A< 5B\Rightarrow A< B\)
Áp dụng công thức : \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(a;b;m\in N\right)\)
Ta có : \(A=\frac{5^{12}+1}{5^{13}+1}< 1\)
\(\Leftrightarrow A=\frac{5^{12}+1}{5^{13}+1}< \frac{5^{12}+1+4}{5^{13}+1+4}=\frac{5^{12}+5}{5^{13}+5}=\frac{5\left(5^{11}+1\right)}{5\left(5^{12}+1\right)}=B\)
\(\Leftrightarrow A< B\)