so sánh :
a. \(\frac{27}{82}\)và \(\frac{26}{75}\)
b. \(\frac{-49}{78}\)và \(\frac{64}{-95}\)
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a/ Ta có:
\(\frac{3}{124}=\frac{30}{1240}\) ; \(\frac{1}{41}=\frac{30}{1230}\) ; \(\frac{5}{207}=\frac{30}{1242}\) ; \(\frac{2}{83}=\frac{30}{1245}\)
Vì các phân số trên đều cùng tử nên ta so sánh mẫu : 1230<1240<1242<1242
=> \(\frac{30}{1230}>\frac{30}{1240}>\frac{30}{1242}>\frac{30}{1245}\)
Hay : \(\frac{1}{41}>\frac{3}{124}>\frac{5}{207}>\frac{2}{83}\)
b/ Ta có:
\(\frac{16}{9}=\frac{48}{27};\frac{24}{13}=\frac{48}{26}\)
Vì 27>26
=> \(\frac{16}{9}< \frac{24}{13}\)
3124=3012403124=301240 ; 141=301230141=301230 ; 5207=3012425207=301242 ; 283=301245283=301245
Vì các phân số trên đều cùng tử nên ta so sánh mẫu : 1230<1240<1242<1242
=> 301230>301240>301242>301245301230>301240>301242>301245
Hay : 141>3124>5207>283141>3124>5207>283
b/ Ta có:
169=4827;2413=4826169=4827;2413=4826
Vì 27>26
=> 169<2413169<2413
Ta có : \(A=\frac{3^{10}+1}{3^9+1}\) => \(A.\frac{1}{3}=\frac{3^{10}+1}{3^{10}+3}=\frac{\left(3^{10}+3\right)-2}{3^{10}+3}=1-\frac{2}{3^{10}+3}\)
\(B.\frac{1}{3}=\frac{3^9+1}{3^8+1}\Rightarrow B.\frac{1}{3}=\frac{3^9+1}{3^9+3}=\frac{\left(3^9+3\right)-2}{3^9+3}=1-\frac{2}{3^9+3}\)
Vì : \(\frac{2}{3^{10}+3}< \frac{2}{3^9+3}\) nên \(A>B\)
\(a,\frac{27}{82}< \frac{27}{83}=\frac{1}{3};\frac{26}{75}>\frac{25}{75}=\frac{1}{3}\)
nên\(\frac{27}{82}< \frac{26}{75}\)
\(b,\frac{49}{78}< \frac{52}{78}=\frac{2}{3};\frac{64}{95}>\frac{64}{96}=\frac{2}{3}\)
nên\(\frac{49}{78}< \frac{64}{95}\Rightarrow\frac{-49}{78}>\frac{64}{-95}\)
c, Rút gọn:\(\frac{2525}{2929}=\frac{25}{29};\frac{217}{245}=\frac{31}{35}\)
Ta có:\(1-\frac{25}{29}=\frac{4}{29};1-\frac{31}{35}=\frac{4}{35}\Rightarrow1-\frac{25}{29}>1-\frac{31}{35}\)
\(\Rightarrow\frac{25}{29}< \frac{31}{35}\)hay\(\frac{2525}{2929}< \frac{217}{245}\)
\(d,A=\frac{3^{10}+1}{3^9+1}=1+\frac{3}{3^9+1}\);\(B=\frac{3^9+1}{3^8+1}=1+\frac{3}{3^8+1}\)
Dễ dàng nhận thấy \(\frac{3}{3^9+1}< \frac{3}{3^8+1}\Rightarrow A< B\)
Xin lỗi bạn e, mk ko làm được. Chúc bạn học tốt
1 Bài làm
a) Ta có :27/ 82 = 0,329...
26 /75= 0,346...
Vì 0,329...<0,346... nên 27/82<26/75
b) Ta có 134/3 = 44,66...
55/21 = 2,61...
74/19= 3,89...
116/37= 3.15.....
Vì 44,66.. > 3,89...>3,15...>2,61 nên 134/3>74/19>116/37>55>21
c) Ta có 16/9 : 24/13 = 0,96.....
mà 0,96<1 nên 16/9<24,13
d) Ta có -49/78= -0,628...
64:-95 = -0,673...
Vì -0,628...> - 0,673... nên -49/78>64:-95
a) 27/82 < 26/75 ( 2025/6250 < 2132\6250)
b) -49/78 > 64/ -95 ( - 3136/7410 > -4992/7410)
c) ta có: \(A=\frac{54.107-53}{53.107}=\frac{53.107+(107-53)}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+\left(269-133\right)}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow A< B\)
d) ta có: \(A=\frac{3^{10}+1}{3^9+1}=\frac{3.\left(3^9+1\right)-2}{3^9+1}=\frac{3.\left(3^9+1\right)}{3^9+1}-\frac{2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3.\left(3^8+1\right)-2}{3^8+1}=\frac{3.\left(3^8+1\right)}{3^8+1}-\frac{2}{3^8+1}=3-\frac{2}{3^8+1}\)
mà \(\frac{2}{3^9+1}< \frac{2}{3^8+1}\Rightarrow3-\frac{2}{3^9+1}< 3-\frac{2}{3^8+1}\)
=> A < B
Ta có -49.(-95) = 4655 < 4992 = 64.78
=> \(\frac{-49}{78}\)< \(\frac{64}{-95}\)
a) Ta có:
+) \(\dfrac{1}{2}=\dfrac{3}{6}\)
+) \(\dfrac{1}{3}=\dfrac{2}{6}\)
+) \(\dfrac{2}{3}=\dfrac{4}{6}\)
=> \(\dfrac{2}{6}< \dfrac{3}{6}< \dfrac{4}{6}\)
hay \(\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)
b) Ta có:
+) \(\dfrac{4}{9}=\dfrac{56}{126}\)
+) \(-\dfrac{1}{2}=-\dfrac{63}{126}\)
+) \(\dfrac{3}{7}=\dfrac{54}{126}\)
=> \(-\dfrac{63}{126}< \dfrac{54}{126}< \dfrac{56}{126}\)
hay \(-\dfrac{1}{2}< \dfrac{3}{7}< \dfrac{4}{9}\)
c) Ta có:
+) \(\dfrac{27}{82}=\dfrac{2025}{6150}\)
+) \(\dfrac{26}{75}=\dfrac{2132}{6150}\)
=> \(\dfrac{2025}{6150}< \dfrac{2132}{6150}\)
hay \(\dfrac{27}{82}< \dfrac{26}{75}\)
d) Ta có:
+) \(-\dfrac{49}{78}=-\dfrac{4655}{7410}\)
+) \(-\dfrac{64}{95}=-\dfrac{4992}{7410}\)
=> \(-\dfrac{4665}{7410}>-\dfrac{4992}{7410}\)
hay \(-\dfrac{49}{78}>-\dfrac{64}{95}\)
giúp mk với
so sanh -49/78