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ta có: \(\frac{1717}{2525}=\frac{17}{25}\)
\(\frac{515151}{727272}=\frac{51}{72}=\frac{17}{24}>\frac{17}{25}\)
\(\Rightarrow\frac{1717}{2525}< \frac{515151}{727272}\)
A)1212/1313<1313/1414.B)1717/2525<515151/727272.23/28,7/9,5/7,11/18
\(\dfrac{121212}{131313}=\dfrac{121212:10101}{131313:10101}=\dfrac{12}{13}=1-\dfrac{1}{13}\)
\(\dfrac{1313}{1414}=\dfrac{1313:101}{1414:101}=\dfrac{13}{14}=1-\dfrac{1}{14}\)
Vì \(\dfrac{1}{13}>\dfrac{1}{14}\) nên \(\dfrac{12}{13}< \dfrac{13}{14}\) hay \(\dfrac{1212}{1313}< \dfrac{1313}{1414}\)
b) \(\dfrac{1717}{2525}=\dfrac{1717:101}{2525:101}=\dfrac{17}{25}=\dfrac{51}{75}\)
\(\dfrac{515151}{727272}=\dfrac{515151:10101}{727272:10101}=\dfrac{51}{72}\)
Vì \(\dfrac{51}{75}< \dfrac{51}{72}\) nên \(\dfrac{17}{25}< \dfrac{51}{72}\) hay \(\dfrac{1717}{2525}< \dfrac{515151}{727272}\)
a) \(\dfrac{1212}{1313}=\dfrac{101x12}{101x13}=\dfrac{12}{13}< \dfrac{12+1}{13+1}=\dfrac{13}{14}\)
\(\dfrac{1313}{1414}=\dfrac{101x13}{101x14}=\dfrac{13}{14}\)
Vậy \(\dfrac{1212}{1313}< \dfrac{1313}{1414}\)
Làm tương tự câu b
Vì 14/14 = 1; 15/15 = 1 ; ....
⇔ Tổng trên là 6 < 11.
Vì 20/20 = 1 ; 21/21 = 1 ; ...
⇒ Tổng trên là 6 < 15.
A = \(\dfrac{1414+1515+1616+1717+1818+1919}{2020+2121+2022+2323+2424+2525}\)
A = \(\dfrac{101\times\left(14+15+16+17+18+19\right)}{101\times\left(20+21+22+23+24+25\right)}\)
A = \(\dfrac{99}{135}\)
A = \(\dfrac{11}{15}\)
1717/1919=17/19
171717/191919=17/19
Vậy 1717/1919 = 171717/191919
\(\frac{1414+1515+1616+1717+1818+1919}{2000+2121+2222+2323+2424+2525}\)
\(=\frac{101.\left(14+15+16+17+18+19\right)}{101.\left(20+21+22+23+24+25\right)}\)
\(\frac{14+15+16+17+18+19}{20+21+22+23+24+25}=\frac{99}{135}=\frac{11}{15}\)
L_I_K_E NHA
\(\frac{1717}{2525}=\frac{17}{25};\frac{515151}{727272}=\frac{51}{72}\)
Ta có : \(\frac{17}{25}
\(\frac{1717}{2525}=\frac{17}{25}\)
\(\frac{515151}{727272}=\frac{17}{24}\)
Cùng tử, ta có : 25 > 24
=> \(\frac{17}{25}