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\(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.71\)
\(72^{44}-72^{43}=72^{43}.\left(72-1\right)=72^{43}.71\)
Vì \(72^{44}>72^{43}\Rightarrow72^{45}-72^{44}>72^{44}-72^{43}\)
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a) Ta có:
\(\frac{15}{301}>\frac{15}{300}=\frac{1}{20}\)
\(\frac{25}{499}< \frac{25}{500}=\frac{1}{20}\)
Vì \(\frac{1}{20}=\frac{1}{20}\) nên \(\frac{15}{301}>\frac{1}{20}>\frac{25}{499}\) hay \(\frac{15}{301}=\frac{25}{499}\)
Vậy \(\frac{15}{301}>\frac{25}{499}\)
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Ta có:\(\frac{111}{332}\)>\(\frac{111}{333}\)=\(\frac{1}{3}\)\(\Rightarrow\)\(\frac{111}{332}\)>\(\frac{1}{3}\)
\(\frac{166}{499}\)<\(\frac{166}{498}\)=\(\frac{1}{3}\)\(\Rightarrow\)\(\frac{166}{499}\)<\(\frac{1}{3}\)
\(\Rightarrow\)\(\frac{111}{332}\)>\(\frac{166}{499}\)
Vậy\(\frac{111}{332}\)>\(\frac{166}{499}\)
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a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)
b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)
c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)
\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)
\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)
Tương tự câu d,e,f bạn tự làm nhé
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Cách 1 :
\(\frac{-60}{-72}\)\(=\frac{60}{72}=\frac{5}{6}\)
\(\frac{-14}{21}=\frac{-2}{3}\)
Vì \(\hept{\begin{cases}6⋮3\\6⋮6\end{cases}}\Rightarrow BCNN\left(6;3\right)=6\Rightarrow MC=6\)
Có: \(\frac{-2}{3}=\frac{\left(-2\right).2}{3.2}=\frac{-4}{6}\)giữ nguyên \(\frac{5}{6}\).
Vì - 4 < 5 nên \(\frac{-4}{6}< \frac{5}{6}\)hay \(\frac{-14}{21}< \frac{-60}{-72}\)
C2 : Thấy : \(\frac{-14}{21}< 0;\frac{-60}{-72}=\frac{60}{72}>0\)
\(\Rightarrow\frac{-14}{21}< \frac{-60}{-72}\)
Cách 1\(\frac{-14}{21}=\frac{-14:7}{21:7}\frac{2}{3}\)
\(\frac{-60}{-72}=\frac{-60:(-12)}{-72:(-12)}=\frac{5}{6}\)
Cách 2 : Tự làm
\(\frac{72}{145}< \frac{72}{144}=\frac{1}{2}\)
\(\frac{250}{499}>\frac{250}{500}=\frac{1}{2}\)
\(\Rightarrow\frac{72}{145}< \frac{250}{499}\)
72/145<250/499