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\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
Ta có: \(\frac{-29}{73}=\frac{-29\cdot49}{73\cdot49}=\frac{-1421}{3577}\)
\(\frac{-80}{49}=\frac{-80\cdot73}{49\cdot73}=\frac{-5840}{3577}\)
Vì \(\frac{-1421}{3577}>\frac{-5840}{3577}\) nên \(-\frac{29}{73}>\frac{-80}{49}\)
Cách 2:
Ta có: \(-\frac{29}{73}>-1\)
\(-\frac{80}{49}< -1\)
Do đó: \(-\frac{80}{49}< -1< \frac{-29}{73}\)
hay \(-\frac{80}{49}< \frac{-29}{73}\)
a ) Ta có
\(\frac{29}{33}>\frac{29}{37}\)( đồng tử khác mẫu )
\(\frac{22}{37}< \frac{29}{37}\)( đồng mẫu khác tử )
=> \(\frac{29}{33}>\frac{29}{37}>\frac{22}{37}\)
b ) \(\frac{163}{257}< \frac{163}{221}\)
\(\frac{162}{257}>\frac{149}{257}\)
\(\Rightarrow\frac{163}{221}>\frac{163}{257}>\frac{149}{257}\)
a) ta có: \(\frac{22}{37}< \frac{29}{37}\)
\(\frac{29}{33}>\frac{29}{37}\)
\(\Rightarrow\frac{22}{37}< \frac{29}{37}< \frac{29}{33}\)
b) ta có: \(\frac{163}{257}>\frac{149}{257}\)
\(\frac{163}{221}>\frac{163}{257}\)
\(\Rightarrow\frac{163}{221}>\frac{163}{257}>\frac{149}{257}\)
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=(198-1)(1998+1)=1998^2-1<B
M=2003^9.(2003+1)=2003^9.2004<N
P=2^50-1<Q
ta có
\(-\frac{80}{49}< -\frac{49}{49}=1=-\frac{73}{73}< -\frac{29}{73}\)