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mình gợi ý nhé 1
1. là so sánh với 1
2 . so sánh với 0
3 . rút gọn đi rồi quy đồng lên sau đó so sánh
Câu 1 :
\(\dfrac{-25}{37}\&\dfrac{-20}{31}\)
Ta thấy \(\dfrac{-25}{37}< \dfrac{-20}{37}\)
mà \(\dfrac{-20}{37}< \dfrac{-20}{31}\)
\(\Rightarrow\dfrac{-25}{37}< \dfrac{-20}{31}\)
Câu 2 :
\(\dfrac{2}{3}\&\dfrac{5}{7}\)
\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)
\(\Rightarrow\dfrac{5}{7}>\dfrac{2}{3}\) Câu 3 : \(\dfrac{8}{13}\&\dfrac{5}{7}\)Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)
\(\Rightarrow\dfrac{8}{13}< \dfrac{5}{7}\)
Ta có:
\(1-\frac{175}{176}=\frac{1}{176}\)
\(1-\frac{2008}{2009}=\frac{1}{2009}\)
Vì \(\frac{1}{2009}< \frac{1}{176}\)
Do đó\(\frac{2008}{2009}< \frac{175}{176}\)
3n + 3 + 3n + 1 + 2n + 3 + 2n + 2
= 3n.33 + 3n.3 + 2n.23 + 2n.22
= 3n.(27 + 3) + 2n.(8 + 4)
= 3n.30 + 2n.12
= 3n.5.6 + 2n.2.6
= 6.(3n.5 + 2n.2) \(⋮\) 6
a. Ta có:
\(\frac{33}{34}=1-\frac{1}{34}\)
\(\frac{34}{35}=1-\frac{1}{35}\)
Do \(\frac{1}{34}>\frac{1}{35}\Rightarrow1-\frac{1}{34}< 1-\frac{1}{35}\)
\(\Leftrightarrow\frac{33}{34}< \frac{34}{35}\)
\(\Rightarrow\frac{-33}{34}>\frac{-34}{35}.\)
Ta có :
\(1-\frac{1}{34}=\frac{33}{34}\)
\(1-\frac{1}{35}=\frac{34}{35}\)
Do \(\frac{1}{34}>\frac{1}{35}\)
\(\Rightarrow1-\frac{1}{34}< 1-\frac{1}{35}\)
\(\Rightarrow\frac{33}{34}< \frac{34}{35}\)
\(\Rightarrow\frac{-33}{34}>-\frac{34}{35}\)
Vậy \(-\frac{33}{34}>-\frac{34}{35}\)
~ Ủng hộ nhé
Ta có: \(\frac{2000}{-2001}=-\frac{2000}{2001}=-\left(\frac{2001-1}{2001}\right)=-\left(\frac{2001}{2001}-\frac{1}{2001}\right)=-\left(1-\frac{1}{2001}\right)=-1+\frac{1}{2001}\)
\(-\frac{2003}{2002}=-\left(\frac{2002+1}{2002}\right)=-\left(\frac{2002}{2002}+\frac{1}{2002}\right)=-\left(1+\frac{1}{2002}\right)=-1-\frac{1}{2002}\)
Vì \(\frac{1}{2001}>-\frac{1}{2002}\) nên \(-1+\frac{1}{2001}>-1-\frac{1}{2002}\)
hay \(\frac{2000}{-2001}>-\frac{2003}{2002}\)
a) ta có: -33/ 37 = -0,89
-34/35 = -0,97
=> -0,89 > -0,97 => -33/37> -34/35
b) ta có: \(\frac{n+1}{n+2}=\frac{n}{n+2}+\frac{1}{n+2}\)
mà \(\frac{n}{n+2}>\frac{n}{n+3}\Rightarrow\frac{n}{n+2}+\frac{1}{n+2}>\frac{n}{n+3}\)
\(\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)
a) ta có: \(\frac{-33}{7}\) = -0,89
\(\frac{-34}{35}\)= -0,97
=> -0,89 > -0,97 => \(\frac{-33}{37}\)> \(\frac{-34}{35}\)
b) ta có: n+1n+2 =nn+2 +1n+2
mà nn+2 >nn+3 ⇒nn+2 +1n+2 >nn+3
⇒n+1n+2 >nn+3