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3)\(\dfrac{-41}{32}\left(\dfrac{15}{8}-\dfrac{16}{41}\right)+\dfrac{15}{8}\left(\dfrac{41}{32}-\dfrac{8}{3}\right)\)
=\(\dfrac{-41}{32}.\dfrac{15}{8}-\dfrac{-41}{32}.\dfrac{16}{41}+\dfrac{15}{8}.\dfrac{41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(\left(\dfrac{-41}{32}.\dfrac{15}{8}+\dfrac{15}{8}.\dfrac{41}{32}\right)+\dfrac{-16}{41}.\dfrac{-41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(0+\dfrac{1}{2}-5=\dfrac{-9}{2}\)
4)\(\dfrac{13}{29}\left(\dfrac{29}{5}-\dfrac{45}{8}\right)-\dfrac{45}{8}\left(\dfrac{9}{8}-\dfrac{13}{29}\right)\)
=\(\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{9}{8}-\dfrac{45}{8}.\dfrac{13}{29}\)
=\(\left(\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{13}{29}\right)-\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{9}{8}\)
=\(0-\dfrac{13}{5}-\dfrac{405}{64}=\dfrac{-2857}{320}\)
a: 51/56=1-5/56
61/66=1-5/66
mà -5/56<-5/66
nên 51/56<61/66
b: 41/43<1<172/165
c: \(\dfrac{101}{506}>0>-\dfrac{707}{3534}\)
a: \(\dfrac{5}{24}< \dfrac{15}{24}=\dfrac{5}{8}\)
b: \(\dfrac{6+9}{6\cdot9}=\dfrac{15}{54}\)
4/9=24/54
2/3=36/54
Do đó: \(\dfrac{15}{54}< \dfrac{24}{54}< \dfrac{36}{54}\)
a)\(\dfrac{1212}{2323}=\dfrac{1212:101}{2323:101}=\dfrac{12}{23}\)
b)\(\dfrac{-3435}{4141}< \dfrac{-3434}{4141}=\dfrac{-3434:101}{4141:101}\)
Nhận xét:
\(\dfrac{\overline{abab}}{\overline{cdcd}}=\dfrac{\overline{ab}}{\overline{cd}}\)
\(=\dfrac{29}{22}+\dfrac{41}{44}+\dfrac{33}{52}+\dfrac{73}{182}\)
\(=\dfrac{58+41}{44}+\dfrac{33}{52}+\dfrac{73}{182}\)
\(=\dfrac{9}{4}+\dfrac{33}{52}+\dfrac{73}{182}=\dfrac{23}{7}\)
a) Giải
So sánh từng số hạng của A với B, ta thấy:
\(\dfrac{19}{41}< \dfrac{21}{41};\dfrac{23}{53}< \dfrac{23}{49}\) và \(\dfrac{29}{61}< \dfrac{33}{65}\) (vì 29.65 < 33.61)
\(\Rightarrow\dfrac{19}{41}+\dfrac{23}{53}+\dfrac{29}{61}< \dfrac{21}{41}+\dfrac{23}{49}+\dfrac{33}{65}\)
\(\Rightarrow A< B\)
Vậy A < B
b) Giải
Ta có: \(C=\dfrac{19^{20}+5}{19^{20}-8}=\dfrac{19^{20}-8+13}{19^{20}-8}=1+\dfrac{13}{19^{20}-8}\)
\(D=\dfrac{19^{21}+6}{19^{21}-7}=\dfrac{19^{21}-7+13}{19^{21}-7}=1+\dfrac{13}{19^{21}-7}\)
Vì \(19^{20}-8< 19^{21}-7\) và \(13>0\)
\(\Rightarrow\dfrac{13}{19^{20}-8}< \dfrac{13}{19^{21}-7}\)
\(\Rightarrow1+\dfrac{13}{19^{20}-8}< 1+\dfrac{13}{19^{21}-7}\)
\(\Rightarrow\) \(C< D\)
Vậy C < D.
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
Ta có : 8.44= 352
41.9= 369
Vậy 8/41 < 9/44