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a) 13/57=13+16/57+16=29/73 ( Ghi nhớ SKG Toán 6)
-=> 13/57 < 29/73
b) 17/42 = 17-4/42-4 = 13/38
=> 17/42 > 13/38
c)7/41 = 7+6/41+6= 13/47
=> 7/41<13/47
a.Vì \(\frac{17}{19}< 1\) và \(\frac{19}{17}>1\)
nên \(\frac{17}{19}< 1< \frac{19}{17}\)
hay \(\frac{17}{19}< \frac{19}{17}\)
b) \(\frac{15}{7}=2\frac{1}{7}\) và \(\frac{25}{12}=2\frac{1}{12}\)
Vì \(2\frac{1}{7}>2\frac{1}{12}\) nên \(\frac{15}{7}>\frac{25}{12}\)
\(A=\frac{54.107-53}{53.107+54}\)
\(\Leftrightarrow A=\frac{53.107+107-53}{53.107+54}\)
\(\Leftrightarrow A=\frac{53.107+54}{53.107+54}\)
\(\Leftrightarrow A=1\)
\(B=\frac{135.269-133}{134.269+135}\)
\(\Leftrightarrow B=\frac{134.269+269-133}{134.269+135}\)
\(\Leftrightarrow B=\frac{134.269+135}{134.269+135}\)
\(\Leftrightarrow B=1\)
Vì 1 = 1 nên A =B
\(a)\) 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 \(\frac{40}{49}>\frac{15}{21}\)
bạn ơi có thể g cho mình ko quy dong hay so sánh phan so 9/10 va 10/11
Quy đồng: \(\frac{n}{n+1}\)= \(\frac{n\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}\)=\(\frac{n^2.2n}{\left(n+1\right)\left(n+2\right)}\)
\(\frac{n+1}{n+2}\)= \(\frac{\left(n+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+2\right)}\)= \(\frac{n^2+2n+1}{\left(n+1\right)\left(n+2\right)}\)
Vì n2+2n+1 < n2.2n+1 nên...
Vậy...
Ko chắc nha
Nghe nó ko có lý kiểu j j ý
a) \(\frac{8}{9}=1-\frac{1}{9}\)
\(\frac{108}{109}=1-\frac{1}{109}\)
Vì \(\frac{1}{9}>\frac{1}{109}\)
Nên \(1-\frac{1}{9}< 1-\frac{1}{109}\)
Vậy \(\frac{8}{9}< \frac{108}{109}\)
b)
\(\frac{97}{100}=\frac{97\cdot99}{100\cdot99}\)
\(\frac{98}{99}=\frac{98\cdot100}{99\cdot100}\)
\(\Rightarrow\frac{97}{100}< \frac{98}{99}\)
c)
\(\frac{19}{18}=1+\frac{1}{18}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{18}>\frac{1}{2016}\)
Vậy \(\frac{19}{18}>\frac{2017}{2016}\)
d)
\(\frac{133}{173}=\frac{130+3}{170+3}=\frac{13+0,3}{17+0,3}\)
Ta có :
\(\frac{a}{b}< \frac{a+x}{b+x}\forall a;b;x>0\)
Vậy \(\frac{13}{17}< \frac{133}{173}\)