\(\frac{10^2+11^2+12^2}{13^2+14^2}=\frac{10^2+11^2+12^2}{\left(10+3\right)^2+\left(11+3\right)^2}=\frac{10^2+11^2+12^2}{10^2+2.3.10+3^2+11^2+2.3.11+3^2}\)

\(=\frac{10^2+11^2+12^2}{10^2+11^2+2.3.21+3^2.2}=\frac{10^2+11^2+12^2}{10^2+11^2+2.3^2\left(7+1\right)}=\frac{10^2+11^2+12^2}{10^2+11^2+2.3^2.8}\)

\(=\frac{10^2+11^2+12^2}{10^2+11^2+4^2.3^2}=\frac{10^2+11^2+12^2}{10^2+11^2+12^2}=1\)

a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)

\(=\left(100+121+144\right)\div\left(169+196\right)\)

\(=365\div365\)

\(=1\)

b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)

\(=1.2.3...8\left(9-1-8\right)\)

\(=1.2.3...8.0\)

\(=0\)

d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)

\(=1152-374-1152-65+374\)

\(=\left(1152-1152\right)-65+\left(374-374\right)\)

\(=0-65+0\)

\(=-65\)

e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)

\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)

\(=13-1+1+1-1-1+1\)

\(=13+0+0+0\)

\(=13\)

chia từng hạng tử cho nhau

a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)

   =1

b)=(1.2.3....8).(9-1-8)

   =(1.2.3....8).0

   =0

mik chỉ giải được zậy thôi.

t mik nha.

S=​​​\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<\frac{4}{10}+\frac{4}{10}+\frac{4}{10}+\frac{4}{10}+\frac{4}{10}\)

                                                                     =\(\frac{4}{10}\cdot5=2=>S<2\)

       S=\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}\)

=\(\frac{3}{15}\cdot5=1=>S>1\)

Vậy 1<S<2

nhớ k với nhé

b) \(\frac{12}{19}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\frac{12}{19}.\frac{19}{12}.\frac{-13}{17}.\frac{17}{13}.\frac{7}{15}=1.\left(-1\right).\frac{7}{15}=\frac{-7}{15}\)

\(-\frac{5}{7}.\frac{2}{11}+-\frac{5}{7}.\frac{9}{14}+\frac{12}{7}=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{14}\right)+\frac{12}{7}=-\frac{5}{7}.\frac{127}{154}+\frac{12}{7}=-\frac{635}{1078}+\frac{12}{7}=\frac{1213}{1078}\)

\(\frac{12}{19}.\frac{7}{15}.-\frac{13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{12}{19}.\frac{19}{12}\right).\left(-\frac{13}{17}.\frac{17}{13}\right).\frac{7}{15}=1.-1.\frac{7}{15}=-\frac{7}{15}\)

\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)

\(=\left(100+121+144\right):\left(169+196\right)\)

\(=\frac{365}{365}=1\)

(10² + 11² + 12²) : (13² + 14²)

= (100 + 121 + 144) : (169 + 196)

= 365 : 365

= 1

Ủng hộ mk nha ^_-

\(\frac{1}{11^2}+\frac{1}{12^2}+\frac{1}{13^2}+\frac{1}{14^2}+...+\frac{1}{100^2}\)

\(=\frac{1}{11.11}+\frac{1}{12.12}+\frac{1}{13.13}+\frac{1}{14.14}+...+\frac{1}{100.100}\)

\(< \frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{99.100}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{10}-\frac{1}{100}\)

Vì \(\frac{1}{100}>0\Rightarrow\frac{1}{10}-\frac{1}{100}< \frac{1}{10}\)

\(\RightarrowĐPCM\)

theo mình tình thi  \(\frac{1}{11^2}+\frac{1}{12^2}+......+\frac{1}{100^2}=0,08521616902\)

mà \(\frac{1}{10}=0,1\)

\(\Rightarrow0,08521515902< 0,1\)

=365 : 365

=1

(10² +11²+12²):(13² +14²)

=(100+121+144):(169+196)

=365:365

=1