\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{97.101}\\ =1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)

Để A là phân số thì

\(n^2-25\ne0\\ \Leftrightarrow n^2\ne25\\ \Leftrightarrow n\ne\pm5\)

\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2009}{2010}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2009}{2010}\)

\(\Leftrightarrow-\dfrac{1}{x+1}=-\dfrac{1}{2010}\)

\(\Leftrightarrow x+1=2010\Leftrightarrow x=2009\)

D

D

D

\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)

\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)

\(\Leftrightarrow2x+8=-x+11\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Nhân chéo ta được\(-6(x+4)=3(x-11)=>-6x-24=3x-33=>6x-3x-24+33=0=>3x+9=0=>3x=-9=>x=-3\)

\(3,8276< \overline{3,8ab5}< 3,836\)

=>\(276< \overline{ab5}< 360\)

=>\(\left(a,b\right)\in\left\{\left(2;8\right);\left(2;9\right);\left(3;0\right);\left(3;1\right);\left(3;2\right);\left(3;3\right);\left(3;4\right);\left(3;5\right)\right\}\)