So sánh :
a)\(\frac{3}{124},\frac{1}{41},\frac{5}{207},\frac{2}{83}\)
b)\(\frac{-2525}{2929}và\frac{-217}{245}\)
c)\(A=\frac{3^{10}+1}{3^9+1}vàB=\frac{3^9+1}{3^8+1}\)
d)\(\frac{27}{82}và\frac{26}{75}\)
Bài 1: So sánh các ps sau:
a)\(\frac{1}{2};\frac{1}{3};\frac{2}{3}\)
b)\(\frac{4}{9};-\frac{1}{2};\frac{3}{7}\)
c)\(\frac{3}{124};\frac{1}{41};\frac{5}{207};\frac{2}{83}\)
d)\(\frac{134}{43};\frac{55}{21};\frac{74}{19};\frac{116}{37}\)
e)\(\frac{16}{9};\frac{24}{13}\)
g)\(\frac{-2525}{2929};\frac{-217}{245}\)
h)\(\frac{27}{82};\frac{26}{75}\)
i)\(\frac{-49}{75};\frac{64}{-95}\)
So sánh:
a) \(\frac{-2525}{2929}\) và \(\frac{-217}{245}\)
b) \(\frac{27}{82}\) và \(\frac{26}{75}\)
c) A= \(\frac{3^{10}+1}{3^9+1}\) và B= \(\frac{3^9+1}{3^8+1}\)
A=\(\frac{3^{10}+1}{3^9+1}\)>1
=> A=\(\frac{3^{10}+1}{3^9+1}\)> \(\frac{3^{10}+1+2}{3^9+1+2}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^{10}+3}{3^9+3}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3\left(3^9+1\right)}{3.\left(3^8+1\right)}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^9+1}{3^8+1}\)=B
vậy A>B
vì 1 là bạn quy đồng lên
hai là giải thich như sau
\(\frac{25}{29}\)= 1- \(\frac{4}{29}\)
\(\frac{31}{35}\)= 1- \(\frac{4}{35}\) mà \(\frac{4}{29}\)> \(\frac{4}{35}\)
nên 1-\(\frac{4}{29}\)<1-\(\frac{4}{35}\)
vậy \(\frac{25}{29}\)<\(\frac{31}{35}\)
a. \(\frac{-2525}{2929}\)= -\(\frac{25.101}{29.101}\)=\(\frac{-25}{29}\)
\(\frac{-217}{245}\)=\(\frac{-7.31}{7.35}\)=\(\frac{-31}{35}\)
vì \(\frac{25}{29}\)>\(\frac{31}{35}\) => \(\frac{-25}{29}\)<\(\frac{-31}{35}\)
b.ta có 27.75=2025
26.82=2132
=> 26.82> 27.75
vậy \(\frac{27}{82}\)<\(\frac{26}{75}\)
so sánh
\(\frac{3}{124};\frac{1}{41};\frac{5}{207};\frac{2}{83}\)
\(\frac{16}{9};\frac{24}{13}\)
\(\frac{27}{82};\frac{26}{75}\)
\(\frac{134}{43};\frac{55}{21};\frac{74}{19};\frac{116}{37}\)
\(-\frac{2525}{2929};-\frac{217}{245}\)
\(-\frac{4}{78};\frac{64}{-95}\)
\(-\frac{49}{78};\frac{64}{-95}\)
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
So sánh:
\(\frac{3}{124},\frac{1}{41},\frac{2}{83},\frac{5}{207}\)
\(\frac{-2525}{2929}v\text{à}\frac{-217}{245}\)
So sánh các phân số:
a)\(\frac{3}{124},\frac{1}{41},\frac{5}{207},\frac{2}{83}\)
b)\(\frac{134}{43},\frac{55}{21},\frac{74}{19},\frac{116}{37}\)
c)\(\frac{-2525}{2929},,\frac{-217}{245}\)
d)\(\frac{27}{28},\frac{26}{75}\)
e)\(\frac{-49}{78}.\frac{64}{-95}\)
So sánh:
a, \(\frac{27}{82}\)với \(\frac{26}{75}\)
b, \(\frac{-49}{78}\)với \(\frac{64}{-95}\)
c, \(\frac{2525}{2929}\)với \(\frac{217}{245}\)
d, A= \(\frac{3^{10}+1}{3^9+1}\) với B= \(\frac{3^9+1}{3^8+1}\)
e, P= \(\frac{5^4.107-53}{5^3.107+54}\) với Q= \(135.\frac{269-133}{134.269+135}\)
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
a)\(\frac{27}{82}< \frac{27}{81}=\frac{1}{3}=\frac{26}{78}< \frac{26}{75}\)
\(\Rightarrow\frac{27}{82}< \frac{26}{75}\)
d)\(\frac{3^{10}+1}{3^9+1}< \frac{3^{10}+1+2}{3^9+1}=\frac{3^{10}+3}{3^9+1}=\frac{3.\left(3^9+1\right)}{3.\left(3^8+1\right)}=\frac{3^9+1}{3^8+1}=B\)
\(\Rightarrow A< B\)
so sánh :
a) \(\frac{3}{124};\frac{1}{41};\frac{5}{207}và\frac{2}{83}\)
b)\(\frac{16}{9}và\frac{24}{13}\)
c)\(\frac{27}{82}và\frac{26}{72}\)
so sánh
\(\frac{1}{2},\frac{1}{3},\frac{2}{3}\) \(\frac{4}{9},-\frac{1}{2},\frac{3}{7}\)
\(\frac{3}{124},\frac{1}{41},\frac{5}{207},\frac{2}{83}\) \(\frac{134}{43},\frac{55}{21},\frac{74}{19},\frac{116}{37}\)
\(\frac{16}{9},\frac{24}{13}\) \(\frac{-2525}{2929},\frac{-217}{245}\)
\(\frac{27}{82},\frac{26}{75}\) \(\frac{-49}{78},\frac{64}{-95}\)
So sánh các phân số sau :
a) \(\frac{1}{2};\frac{1}{3};\frac{2}{3}\) e) \(\frac{4}{9};-\frac{1}{2};\frac{3}{7}\)
b) \(\frac{3}{124};\frac{1}{41};\frac{5}{207};\frac{2}{83}\) f) \(\frac{134}{43};\frac{55}{21};\frac{74}{19};\frac{116}{37}\)
c)\(\frac{16}{9}\)và \(\frac{24}{13}\) g) \(\frac{-2525}{2929}\)và \(\frac{-217}{245}\)
d) \(\frac{27}{82}\)và \(\frac{26}{75}\) h) \(\frac{-49}{78}\)và \(\frac{64}{-95}\)
a. Ta có :
\(\frac{1}{2}< \frac{1}{3}\)
\(\frac{2}{3}>\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}< \frac{1}{2}< \frac{2}{3}\)