so sánh
\(\dfrac{9}{10}\)và\(\dfrac{13}{15}\)
Hãy so sánh các phân số sau bằng phương pháp so sánh phần bù :
a)\(\dfrac{10}{11}và\dfrac{19}{20}\)
b) \(\dfrac{13}{15}và\dfrac{15}{17}\)
c) \(\dfrac{31}{35}và\dfrac{35}{37}\)
Rút gọn rồi so sánh hai phân số:
a) \(\dfrac{6}{14}\) và \(\dfrac{4}{7}\) b) \(\dfrac{3}{5}\) và \(\dfrac{6}{15}\) c) \(\dfrac{10}{18}\) và \(\dfrac{2}{9}\)
a) \(\dfrac{6}{14}=\dfrac{6:2}{14:2}=\dfrac{3}{7}\)
\(\dfrac{3}{7}< \dfrac{4}{7}\)
b) \(\dfrac{6}{15}=\dfrac{6:3}{15:3}=\dfrac{2}{5}\)
\(\dfrac{3}{5}>\dfrac{2}{5}\)
c) \(\dfrac{10}{18}=\dfrac{10:2}{18:2}=\dfrac{5}{9}\)
\(\dfrac{5}{9}>\dfrac{2}{9}\)
A=\(\dfrac{13^{15}+1}{13^{16}+1}\) và B= \(\dfrac{13^{16}+1}{13^{17}+1}\)
so sánh A và B
\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)
B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)
vậy A=B
\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)
ta có
\(\dfrac{13^{16}+1}{13^{17}+1}< 1\Rightarrow\dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)
vậy B<A
\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)
ta có B<1 nên
\(\dfrac{13^{16}+1}{13^{17}+1}< \dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)
Vậy B<A
Cho N=\(\dfrac{-7}{10^{2015}}\)+\(\dfrac{-15}{10^{2006}}\)và M=\(\dfrac{-15}{10^{2005}}\)+\(\dfrac{-7}{10^{2006}}\)
So sánh M và N (heo mì) TvT
Ta có :
\(N=\dfrac{-7}{10^{2005}}+\dfrac{-15}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}+\dfrac{-8}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2006}}\)
\(M=\dfrac{-15}{10^{2005}}+\dfrac{-7}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-8}{10^{2005}}+\dfrac{-7}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2005}}\)
Lại có :
\(-\dfrac{8}{10^{2006}}>\dfrac{-8}{10^{2005}}\Leftrightarrow M>N\)
Sử dụng tính chất bắc cầu để so sánh các phân số sau:
a) \(\dfrac{1997}{1996}và\dfrac{1996}{1997}\)
b) \(\dfrac{3}{5}và\dfrac{15}{13}\)
\(a,\dfrac{1997}{1996}>1>\dfrac{1996}{1997}\\ b,\dfrac{3}{5}< 1< \dfrac{15}{13}\)
so sánh hai phân số:( \(\dfrac{1}{243}\))9 và (\(\dfrac{1}{83}\))13
`(1/243)^9 = [1/(3^5)]^9 = [(1/3)^5]^9=(1/3)^13`
Vì: `1/3 > 1/83`
`=> (1/3)^13 > 1/(83)^13`.
Cho \(A=\dfrac{13}{25}+\dfrac{9}{10}-\dfrac{11}{15}+\dfrac{13}{21}-\dfrac{15}{28}+\dfrac{17}{36}-...+\dfrac{197}{4851}-\dfrac{199}{4950}\)
Chứng minh \(A>\dfrac{9}{10}\)
2/ So sánh các phân số sau :
a/ \(\dfrac{7}{10}\) và \(\dfrac{11}{15}\) ; b/ \(\dfrac{-1}{8}\) và \(\dfrac{-5}{24}\) ; c/ \(\dfrac{25}{100}\) và \(\dfrac{10}{40}\)
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
a)
b)
c) \(\dfrac{25}{100}=\dfrac{10}{40}\)
so sánh các phân số sau:
a)\(\dfrac{-8}{9}\) và \(\dfrac{-7}{9}\)
b)\(\dfrac{6}{7}\) và \(\dfrac{11}{10}\)
a)\(\dfrac{-8}{9}< \dfrac{-7}{9}\\ \dfrac{6}{7}< \dfrac{11}{10}\)
a) Vì -8<-7 nên \(\dfrac{-8}{9}< \dfrac{-7}{9}\)
b) Ta có: \(\dfrac{6}{7}< 1;\dfrac{11}{10}>1\)
nên \(\dfrac{6}{7}< \dfrac{11}{10}\)
Giải:
a) -8/9 < -7/9
b) Ta có: 6/7 = 60/70
11/10 = 77/70
Vì 60/70 < 77/70 nên :
6/7 < 11/10