Chứng tỏ rằng: 1/1.2 + 1/2.3 +...+1/9.10 <1
Giải thích
a)11.2+12.3+...+19.10=910<1.
b)32.5+35.8+38.11+311.14+314.17+317.20=12−120<12.
c)Ta có 122 <11.2;132<12.3;142<13.4;...;1252<124.25.
Do đó,122 +132+142+1252<11.2+12.3+13.4+...+124.25<1.
a)11.2+12.3+...+19.10=910<1.
b)32.5+35.8+38.11+311.14+314.17+317.20=12−120<12.
c)Ta có 122 <11.2;132<12.3;142<13.4;...;1252<124.25.
Do đó,122 +132+142+1252<11.2+12.3+13.4+...+124.25<1.