Question

[Hoàng Giang]

Bài 1: Tìm x

a, (x-20)\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{200}}{\dfrac{1}{199}+\dfrac{2}{198}+....+\dfrac{199}{1}}\)= \(\dfrac{1}{2000}\)

b, 2x+\(\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}\)= 10

c, \(\dfrac{3}{35}+\dfrac{3}{63}+\dfrac{3}{99}+....+\dfrac{3}{x\left(x+2\right)}\)= \(\dfrac{24}{35}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{199}{1}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}}{\left(\dfrac{1}{199}+1\right)+\left(\dfrac{2}{198}+1\right)+...+\left(\dfrac{198}{2}+1\right)+1}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}}{\dfrac{200}{2}+\dfrac{200}{3}+...+\dfrac{200}{199}+\dfrac{200}{200}}=\dfrac{1}{200}\)

\(\left(x-20\right)\cdot\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}}{\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{199}{1}}=\dfrac{1}{2000}\)

=>\(\left(x-20\right)\cdot\dfrac{1}{200}=\dfrac{1}{2000}\)

=>\(x-20=\dfrac{1}{10}\)

=>x=20,1

b: \(\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}\)

\(=\left(1+1+1+1+1+1+1+1\right)+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\right)\)

\(=8+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=8+\dfrac{1}{2}-\dfrac{1}{10}=8+\dfrac{4}{10}=8.4\)

\(2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)

=>2x+8,4=10

=>2x=1,6

=>x=0,8

c: \(\dfrac{3}{35}+\dfrac{3}{63}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)

=>\(\dfrac{3}{2}\left(\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)

=>\(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{24}{35}:\dfrac{3}{2}=\dfrac{24}{35}\cdot\dfrac{2}{3}=\dfrac{48}{105}=\dfrac{16}{35}\)

=>\(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{16}{35}\)
=>\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)

=>\(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}=\dfrac{7}{35}-\dfrac{16}{35}=-\dfrac{9}{35}\)

=>x+2=-35/9

=>x=-53/9