(-1/4)^40 > (-1/5)^36

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Bài 1:

a) \(\dfrac{-17}{36}\) và \(\dfrac{23}{-48}\) 

\(\dfrac{-17}{36}=\dfrac{-17.4}{36.4}=\dfrac{-68}{144}\) 

\(\dfrac{23}{-48}=\dfrac{-23}{48}=\dfrac{-23.3}{144.3}=\dfrac{-69}{144}\) 

Vì \(\dfrac{-68}{144}>\dfrac{-69}{144}\) nên \(\dfrac{-17}{36}>\dfrac{23}{-48}\) 

b) \(\dfrac{-1}{3}\) và \(\dfrac{2}{5}\) 

Vì \(\dfrac{-1}{3}\) là số âm mà \(\dfrac{2}{5}\) là số dương nên \(\dfrac{-1}{3}< \dfrac{2}{5}\) 

c) \(\dfrac{2}{7}\) và \(\dfrac{5}{4}\) 

Vì \(\dfrac{2}{7}< 1\) mà \(\dfrac{5}{4}>1\) nên \(\dfrac{2}{7}< \dfrac{5}{4}\) 

d) \(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\) 

\(\dfrac{267}{-268}=\dfrac{-267}{268}=\dfrac{-267.449}{268.449}=\dfrac{-119883}{120332}\) 

\(\dfrac{-1347}{1343}=\dfrac{-1347.89}{1343.89}=\dfrac{-119883}{119527}\) 

Vì \(\dfrac{-119883}{120332}>\dfrac{-119883}{119527}\) nên \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

Bài 2:

\(\dfrac{5}{2}-\left(1\dfrac{3}{7}-0,4\right)=\dfrac{5}{2}-\dfrac{10}{7}-\dfrac{2}{5}=\dfrac{47}{70}\) 

a)

\(\dfrac{48}{92}=\dfrac{48:4}{92:4}=\dfrac{12}{23}\)
\(\dfrac{36}{69}=\dfrac{36:3}{69:3}=\dfrac{12}{23}\)
Ta có: 
Mẫu số chung 2 phân số: 23
Vì \(12=12\) nên \(\dfrac{12}{23}=\dfrac{12}{23}\)
Vậy \(\dfrac{48}{92}=\dfrac{36}{69}\)

b)

\(\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{7}{8}=\dfrac{573}{280}\)

Ta có: 
Mẫu số chung 2 phân số: 280
\(\dfrac{3}{1}=\dfrac{3*280}{1*280}=\dfrac{840}{280}\)
Vì \(573< 840\) nên \(\dfrac{573}{280}< \dfrac{840}{280}\)
Vậy \(\dfrac{573}{280}< \dfrac{3}{1}\)

c)
Ta có: 
Mẫu số chung 2 phân số: 10
\(\dfrac{2}{5}=\dfrac{2*2}{5*2}=\dfrac{4}{10}\)
Vì \(1< 4\) nên\(\dfrac{1}{10}< \dfrac{4}{10}\)
Vậy \(\dfrac{1}{10}< \dfrac{2}{5}\)

d)
\(\dfrac{4}{10}=\dfrac{4:2}{10:2}=\dfrac{2}{5}\)
Ta có: 
Mẫu số chung 2 phân số: 5
Vì \(2=2\) nên \(\dfrac{2}{5}=\dfrac{2}{5}\)
Vậy \(\dfrac{4}{10}=\dfrac{2}{5}\)

Hãy quy đồng hộ mình và đánh dấu nha 

Câu 1: 3^23  >    5^12

Câu 2: 3^36 < 2^8.11^4

Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)

Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)

\(\dfrac{1}{32}>\dfrac{1}{40}\)

\(\dfrac{1}{33}>\dfrac{1}{40}\)

\(\dfrac{1}{34}>\dfrac{1}{40}\)

\(\dfrac{1}{35}>\dfrac{1}{40}\)

\(\dfrac{1}{36}>\dfrac{1}{40}\)

\(\dfrac{1}{37}>\dfrac{1}{40}\)

\(\dfrac{1}{38}>\dfrac{1}{40}\)

\(\dfrac{1}{39}>\dfrac{1}{40}\)

\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)

Vậy \(S>\dfrac{1}{4}\)

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