\(\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}-\dfrac{2}{7}\right)\\ =\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{5}+\dfrac{2}{7}\\ =\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\\ =\dfrac{1}{3}+0+0=\dfrac{1}{3}\)

Đề bài yêu cầu gì hả bạn.

tính kết qủa á

   -49 + 118 - 52

= -(49 + 52) + 118

= -101 + 118

= 17

Ta có:

\(\dfrac{1}{430}\)+\(\dfrac{1}{324}\)

=\(\dfrac{162}{69660}\)+\(\dfrac{215}{69660}\)

=\(\dfrac{162+215}{69660}\)

=\(\dfrac{377}{69660}\)

\(\dfrac{1}{430}+\dfrac{1}{324}\)

\(=\dfrac{162}{69660}+\dfrac{215}{69660}\)

\(=\dfrac{377}{69660}\)

\(\dfrac{1}{555}+\dfrac{1}{678}\)

\(=\dfrac{678}{376290}+\dfrac{555}{376290}\)

\(=\dfrac{1233}{376290}=\dfrac{411}{125430}=\dfrac{137}{41810}\)

Ta có:

\(\dfrac{1}{555}\)+\(\dfrac{1}{678}\)

=\(\dfrac{678}{376290}\)+\(\dfrac{555}{376290}\)

=\(\dfrac{678+555}{376290}\)

=\(\dfrac{1233}{376290}\)

=\(\dfrac{137}{41810}\)

Ta thấy: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\)

\(\dots\)

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

Suy ra: \(A=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dots+\dfrac{1}{100^2}\)

\(< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dots+\dfrac{1}{99\cdot100}\)

\(=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=2-\dfrac{1}{100}< 2\)

Vậy \(A< 2\)

Do x + 2 là ước của 2x - 1 nên (2x - 1) ⋮ (x + 2)

Ta có:

2x - 1 = 2x + 4 - 5 = 2(x + 2) - 5

Để (2x - 1) ⋮ (x + 2) thì 5 ⋮ (x + 2)

⇒ x + 2 ∈ Ư(5) = {-5; -1; 1; 5}

⇒ x ∈ {-7; -3; -1; 3}

\(\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)

\(=\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+\left(1-\dfrac{1}{20}\right)+\left(1-\dfrac{1}{30}\right)+\left(1-\dfrac{1}{42}\right)+\left(1-\dfrac{1}{56}\right)+\left(1-\dfrac{1}{72}\right)+\left(1-\dfrac{1}{90}\right)\)

\(=8-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=8-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=8-\left(\dfrac{3-2}{2\cdot3}+\dfrac{4-3}{3\cdot4}+\dfrac{5-4}{4\cdot5}+\dfrac{6-5}{5\cdot6}+\dfrac{7-6}{6\cdot7}+\dfrac{8-7}{7\cdot8}+\dfrac{9-8}{8\cdot9}+\dfrac{10-9}{9\cdot10}\right)\)

\(=8-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=8-\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=8-\dfrac{4}{10}\)

\(=\dfrac{80}{10}-\dfrac{4}{10}=\dfrac{76}{10}=\dfrac{38}{5}\)

Ta có công thức tổng quát của dãy số: 

\(1\cdot2+2\cdot3+3\cdot4+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

Trong đề bài ta có dãy số: \(1\cdot2+2\cdot3+...+99\cdot100\) có \(n=99\)

\(\Rightarrow1\cdot2+2\cdot3+...+99\cdot100=\dfrac{99\cdot\left(99+1\right)\cdot\left(99+2\right)}{3}=333300\)

Trở lại để bài:

\(\dfrac{1\cdot2+2\cdot3+3\cdot4+...+99\cdot100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+...+\left(x^2+99\right)}=50\dfrac{116}{131}\)

\(\Rightarrow\dfrac{333300}{x^2+x^2+1+x^2+2+...+x^2+99}=\dfrac{6666}{131}\)

\(\Rightarrow\dfrac{333300}{\left(x^2+x^2+...+x^2\right)+\left(1+2+3+...+99\right)}=\dfrac{6666}{131}\)

\(\Rightarrow\dfrac{333300}{100x^2+4950}=\dfrac{6666}{131}\)

\(\Rightarrow6666\left(100x^2+4950\right)=333300\cdot131\)

\(\Rightarrow666600x^2+32996700=43662300\)

\(\Rightarrow666600x^2=10665600\)

\(\Rightarrow x^2=\dfrac{10665600}{666600}\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow x=\pm4\)

Ta có: 1+2-3-4+5+6-7-8+9+...+994-995-996+997+998

= (1-3+5-7+...+993-995+997) + (2-4+6-8+...994-996+998)

= (-2-2-2-2...-2+997) + (-2-2-2...-2)

= 499 + 500

= 999