Sai vì có cung MS là 21 và tử số thì -15<-14 nên phải là : <

sai vì -15<-14

\(\frac{1010+1111+1212+1313+1414+1515+1616+1717}{2020+2121+2222+2323+2424+2525+2626+2727}\)

\(=\frac{101.10+101.11+...+101.17}{101.20+101.21+...+101.27}\)

\(=\frac{101.\left(10+11+...+17\right)}{101.\left(20+21+...+27\right)}\)

\(=\frac{108}{188}\)

\(=\frac{27}{47}\)

\(2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right)\cdot5.y>\frac{5}{6}\)

\(\Rightarrow2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right):5.y>\frac{5}{6}\)

\(\Rightarrow2>\left(\frac{20}{120}+\frac{16}{120}+\frac{9}{120}+\frac{5}{120}\right):5.y>\frac{5}{6}\)

\(\Rightarrow2>\frac{5}{12}:5.y>\frac{5}{6}\)

\(\Rightarrow2>\frac{1}{12}.y>\frac{5}{6}\)

Đặt :\(\frac{1}{12}.y=2\Rightarrow y=2:\frac{1}{12}=24\)

\(\frac{1}{12}.y=\frac{5}{6}\Rightarrow y=\frac{5}{6}:\frac{1}{12}=10\)

\(\Rightarrow24>y>10\)

\(\Rightarrow y\in\left\{11;12;...;23\right\}\)

ko biết

I am sorry

Kb nhé

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

   \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\frac{14}{15}\)

\(=\frac{7}{15}\)

a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{15}\right)=\frac{1}{2}.\frac{14}{15}\)\(=\frac{7}{15}\)

b)\(\frac{1414+1515+...+1919}{2020+2121+...+2525}\)

\(\Rightarrow\frac{101\left(14+15+16+17+18+19\right)}{101\left(20+21+22+23+24+25\right)}\)

\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}\)

\(=\frac{\left(19+14\right).6:2}{\left(25+20\right).6:2}=\frac{19+14}{25+20}=\frac{33}{45}=\frac{11}{15}\)

Đây là những bài lớp 5+ vậy nên không làm theo kiểu lớp 5 được, cố gắng tìm bài đúng lớp nhé bạn (VNNLL cc)

\(\left(\frac{1313}{1414}+\frac{10}{160}\right)-\left(\frac{130}{140}-\frac{1515}{1616}\right)\)

\(=\left(\frac{13}{14}+\frac{1}{16}\right)-\left(\frac{13}{14}-\frac{15}{16}\right)\)

\(=\frac{13}{14}+\frac{1}{16}-\frac{13}{14}+\frac{15}{16}\)

\(=\left(\frac{13}{14}-\frac{13}{14}\right)+\left(\frac{1}{16}+\frac{15}{16}\right)\)

\(=0+1\)

\(=1\)

Có (\(\frac{1313}{1414}\)+\(\frac{10}{160}\)) - (\(\frac{130}{140}\)-\(\frac{1515}{1616}\))

=\(\frac{13}{14}\)+\(\frac{1}{16}\)-\(\frac{13}{14}\)+\(\frac{15}{16}\)

=(\(\frac{13}{14}-\frac{13}{14}\)) + (\(\frac{1}{16}+\frac{15}{16}\))

=0+1=1

Để mình giúp bạn nha !!!

A) \(\frac{7}{15}\)

B) 4,391

\(\frac{1414+1515+1616+1717+1818+1919}{2020+2121+2222+2323+2424+2525}=\frac{9999}{13635}=\frac{11}{15}\)

Ai làm đúng và nhanh nhất mk k cho

cái thứ hai lớn hơn

Ta có : \(\frac{1212+1313+1414+1515+1616}{1717+1818+1919+2020+2121}=\frac{12+13+14+15+16}{17+18+19+20+21}=\frac{28\cdot\frac{5}{2}}{38\cdot\frac{5}{2}}=\frac{14}{19}\)

Dễ thấy \(\frac{111111}{666665}< \frac{1}{2}\) ( do 111111 < 666665/2 ) và \(\frac{14}{19}>\frac{1}{2}\) ( do 14 > 19/2 )

Vậy \(\frac{111111}{666665}< \frac{1212+1313+1414+1515+1616}{1717+1818+1919+2020+2121}\)