\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{200}\)

\(\frac{1}{x+1}=\frac{-299}{200}\)

\(x+1=\frac{-200}{299}\)

\(x=\frac{-499}{299}\)

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\left(\text{​Quy đồng nhé !}\right)\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{800}\)

=> x + 1 = 800 

<=> x = 799

1/3+1/6+1/10+...+1/x(x+1):2=399/400

2.[1/3.2+1/6.2+1/10.2+...+1/x(x+1)]=399/400

2.[1/6+1/12+1/20+...+1/x(x+1)]=399/400

2.[1/2.3+1/3.4+1/4.5+...+1/x(x+1)]=399/400

1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=399/800

1/2-1/x+1=399/800

1/x+1=1/800

=> x+1=800

=> x=799

Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)

=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)

=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)

=> x = 9

Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)

=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)

=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)

=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)

=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)

=> \(\frac{15}{16}:x=\frac{11}{12}\)

=> \(x=\frac{45}{44}\)

Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)

=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)

=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)

=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)

=> \(\frac{1}{x+1}=\frac{1}{800}\)

=> x = 799

Bài 2 :

\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)

Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)

Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)

\(=1-\frac{1}{12}=\frac{11}{12}\) (2)

Thay (1) và (2) vào biểu thức (*) ta được :

\(\frac{15}{16}:x=\frac{11}{12}\)

\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)

\(\Leftrightarrow x=\frac{45}{44}\)

Vậy : \(x=\frac{45}{44}\)

tôi ko biết

=1+(2-3-4+5)+(6-7-8+9)+...+(398-399-400+401)+402

=(2+5-3-4)+(6+9-7-8)+...+(398+401-399-400)+1+402

=0+0+...+0+403

=403

h mình nha

gfdgdfyrdgd