11/32=0,34375;16/49=0.3265306122

vay 11/32 > 16/49

58/89=0,6516853933;36/53=0,679246283

vay 58/89 < 36/53

11/32<16/49

58/89>36/53

a) <

b)<

c)<

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)A=54-53/53+54=1/107=2/214

B=135-133/134+135=2/169

tự so sánh tiếp

Đm giải nốt câu b đi bạn

\(a)\) Ta có : 

\(\frac{51}{85}=\frac{3}{5}\)

\(\frac{58}{145}=\frac{2}{5}\)

Vì \(\frac{3}{5}>\frac{2}{5}\) nên \(\frac{51}{85}>\frac{58}{145}\)

Vậy \(\frac{51}{85}>\frac{58}{145}\)

\(b)\) Ta có : 

\(\frac{69}{-230}=\frac{-3}{10}\)

\(\frac{-39}{143}=\frac{-3}{11}\)

Vì \(\frac{-3}{10}< \frac{-3}{11}\) nên \(\frac{69}{-230}< \frac{-39}{143}\)

Vậy \(\frac{69}{-230}< \frac{-39}{143}\)

\(c)\) Ta có : 

\(1+\frac{-7}{41}=\frac{34}{41}\)

\(1+\frac{13}{-47}=\frac{34}{47}\)

Vì \(\frac{34}{41}>\frac{34}{47}\) nên \(1+\frac{-7}{41}>1+\frac{13}{-47}\) hay \(\frac{-7}{41}>\frac{13}{-47}\)

Vậy \(\frac{-7}{41}>\frac{13}{-47}\)

\(d)\) Ta có : 

\(1-\frac{40}{49}=\frac{9}{49}\)

\(\frac{15}{21}=\frac{5}{7}=\frac{35}{49}< \frac{40}{49}\)

Vậy \(\frac{40}{49}>\frac{15}{21}\)

\(\frac{12}{47}>\frac{12}{48}=\frac{1}{4}\)

\(\frac{19}{77}< \frac{19}{76}=\frac{1}{4}\)

\(\Rightarrow\frac{12}{47}>\frac{19}{77}\)

\(\frac{58}{89}< \frac{58}{87}=\frac{2}{3}\)

\(\frac{36}{53}>\frac{36}{54}=\frac{2}{3}\)

\(\Rightarrow\frac{58}{89}< \frac{36}{53}\)

a) 11/32>16/49

b)456/461>123/128

c)12/47>19/77 

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