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Ta có :
\(1-\frac{38}{133}=\frac{133}{133}-\frac{38}{133}=\frac{95}{133}=\frac{5}{7}\)
\(1-\frac{129}{344}=\frac{344}{344}-\frac{129}{344}=\frac{215}{344}=\frac{5}{8}\)
Vì \(\frac{5}{7}>\frac{5}{8}\) nên \(\frac{38}{133}< \frac{129}{344}\)
\(\Rightarrow\)\(\frac{38}{-133}>\frac{-129}{344}\)
Vậy \(\frac{38}{-133}>\frac{-129}{344}\)
38/133>38/344(vì 133<344).
Mà 38/344>29/344(vì 38>29).
=>38/133>29/344.
tk mk nha các bn.
-chúc ai tk cho mk học giỏi và may mắn ,thanks các bn nhìu-
\(A=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(B=\frac{135\cdot268-133}{134\cdot269+135}=\frac{\left(134+1\right)\cdot268-133}{134\cdot269+135}=\frac{134\cdot268+268-133}{34\cdot269+135}=\frac{134\cdot268+135}{134\cdot269+135}=1\)
Vì 1=1 nên A=B
A= (54.107-53)/(53.107+54)
= (53+1).107-53 / 53.107+54
=53.107+107-53 / 53.107+54
=53.107+54 / 54.107 + 54
=1
B= 135.269-133 / 134.269+135
= (134+1).269-133 / 134.269+135
= 134.269+269-133 / 134.269+135
=134.269+136 / 134.269+135
=134.269+135/ 134.269+135 + 1/134.269+135
=1 + 1/134.269+135 >1=A
\(A=\frac{54.107-53}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}>1\)
\(A=\frac{54.107-53}{53.107+54}<\frac{135.269-133}{134.269+135}\)
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 \(\frac{13}{17}< \frac{133}{173}\)