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\(\frac{2323+1313+1414+1515+1616}{4141+2222+2323+2424+2525}\)
=\(\frac{8181}{13635}\)
\(\dfrac{121212}{131313}=\dfrac{121212:10101}{131313:10101}=\dfrac{12}{13}=1-\dfrac{1}{13}\)
\(\dfrac{1313}{1414}=\dfrac{1313:101}{1414:101}=\dfrac{13}{14}=1-\dfrac{1}{14}\)
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
\(\frac{1010+1111+1212+1313+1414+1515}{2020+2121+2222+2323+2424+2525}=\frac{10\times101+11\times101+12\times101+13\times101+14\times101+15\times101}{20\times101+21\times101+22\times101+23\times101+24\times101+25\times101}=\frac{101\times\left(10+11+12+13+14+15\right)}{101\times\left(20+21+22+23+24+25\right)}\)
\(=\frac{10+11+12+13+14+15}{20+21+22+23+24+25}=\frac{75}{135}=\frac{5}{9}\)
1010+1111+1212+1313+1414+1515= 7575
2020+2121+2222+2323+2424+2525= 13635
A)1212/1313<1313/1414.B)1717/2525<515151/727272.23/28,7/9,5/7,11/18
a) \(\dfrac{12}{14}=\dfrac{1200}{1400}=\dfrac{1400-200}{1400}=1-\dfrac{200}{1400}\)
\(\dfrac{1212}{1414}=\dfrac{1414-200}{1414}=1-\dfrac{200}{1414}\)
vì \(\dfrac{200}{1414}< \dfrac{200}{1400}\)
Nên \(1-\dfrac{200}{1400}< 1-\dfrac{200}{1414}\)
Vậy \(\dfrac{12}{14}< \dfrac{1212}{1414}\)
Các bài sau tương tự
44141
1313+41414+1414
=42727+1414
=44141