Lời giải:

Ta có:

\(\frac{1}{13}; \frac{1}{14}; \frac{1}{15}<\frac{1}{12}\)

\(\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{61}; \frac{1}{62};\frac{1}{63}< \frac{1}{60}\)

\(\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}\)

Do đó:

\(A< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{9}{20}+\frac{1}{20}\)

\(\Leftrightarrow A< \frac{1}{2}\) (đpcm)

Đặt biểu thức bằng A:

\(\Rightarrow A=\dfrac{1}{5}\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Ta thấy: \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{61}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{3}{31}+\dfrac{3}{61}< \dfrac{1}{2}\left(đpcm\right)\)

\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)

\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)

\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)

a: \(\left(5+\frac15-\frac29\right)-\left(2-\frac{1}{23}-\frac{3}{35}+\frac56\right)-\left(8+\frac27-\frac{1}{18}\right)\)

\(=5+\frac15-\frac29-2+\frac{1}{23}+\frac{3}{35}-\frac56-8-\frac27+\frac{1}{18}\)

\(=\left(5-2-8\right)+\left(\frac15+\frac{3}{35}-\frac27\right)+\left(-\frac29-\frac56+\frac{1}{18}\right)+\frac{1}{23}\)

\(=\left(-5\right)+\left(\frac{7}{35}+\frac{3}{35}-\frac{10}{35}\right)+\left(-\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)

\(=-5+\left(-\frac{18}{18}\right)+\frac{1}{23}=-6+\frac{1}{23}=-\frac{138}{23}+\frac{1}{23}=-\frac{137}{23}\)

c: \(-\frac57-\left(-\frac{5}{67}\right)+\frac{13}{10}+\frac12+\left(-\frac16\right)+1\frac{3}{14}-\left(-\frac25\right)\)

\(=-\frac57+\frac{5}{67}+\frac{13}{10}+\frac12-\frac16+\frac{17}{14}+\frac25\)

\(=\left(-\frac57+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac25-\frac16\right)+\frac{5}{67}\)

\(=\left(-\frac{10}{14}+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac{4}{10}-\frac16\right)+\frac{5}{67}\)

\(=\left(\frac{7}{14}+\frac12\right)+\left(\frac{17}{10}-\frac16\right)+\frac{5}{67}=1+\frac{5}{67}+\frac{51}{30}-\frac{5}{30}\)

\(=\frac{72}{67}+\frac{46}{30}=\frac{72}{67}+\frac{23}{15}=\frac{2621}{1005}\)

d: \(\frac35:\left(-\frac{1}{15}-\frac16\right)+\frac35:\left(-\frac13-1\frac{1}{15}\right)\)

\(=\frac35:\left(-\frac{2}{30}-\frac{5}{30}\right)+\frac35:\left(-\frac{5}{15}-\frac{16}{15}\right)\)

\(=\frac35:\left(-\frac{7}{30}\right)+\frac35:\left(-\frac{21}{15}\right)\)

\(=\frac35\cdot\frac{-30}{7}+\frac35\cdot\frac{-5}{7}=\frac35\cdot\left(-\frac{30}{7}-\frac57\right)\)

\(=\frac35\cdot\left(-\frac{35}{7}\right)=\frac35\cdot\left(-5\right)=-3\)

chịu thui

chúc bn học gioi!

nhaE@@

Toán lớp 7        bye mk đi hc đây hihi

 $$$$   

Ta có : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)

= \(\frac{1}{3}+\)( \(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\)) \(+\)( \(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\) ) \(< \)\(\frac{1}{3}+\)( \(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\)) \(+\)( \(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\)) = \(\frac{1}{2}\)

Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)

\(A=1-\frac{2}{3}+1-\frac{2}{15}+1-\frac{2}{35}+1-\frac{2}{63}+1-\frac{2}{99}+1-\frac{2}{143}\)      

\(=1+1+1+1+1+1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}-\frac{2}{143}\)   

\(=6-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)   

\(=6-\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)   

\(=6-\left(1-\frac{1}{13}\right)\)   

\(=6-1+\frac{1}{13}\)   

\(=5+\frac{1}{13}\)   

\(=\frac{65}{13}+\frac{1}{13}\)   

\(=\frac{66}{13}\)