=335510

335510

\(B=1+\frac{3}{5}+\frac{3}{35}+\frac{3}{55}+.....+\frac{3}{9999}\)

\(B=\frac{3}{1}.3+\frac{3}{3}.5+\frac{3}{5}.7+......+\frac{3}{99}.101\)

\(B=\frac{3}{2}\left(\frac{2}{1}.3+\frac{2}{3}.5+.......+\frac{3}{99}.101\right)\)

\(B=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{101}\right)\)

\(B=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(B=\frac{3}{2}.\frac{100}{101}\)

\(B=\frac{150}{101}\)

#)Giải :

\(P=1+\frac{9}{45}+\frac{9}{105}+\frac{9}{189}+...+\frac{9}{29997}\)

\(P=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)

\(P=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101} \right)\)

\(P=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(P=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(P=\frac{3}{2}\times\frac{100}{101}\)

\(P=\frac{150}{101}\)

trả lời 

=150/101 

chúc bn 

hc tốt

trả lời 

=150/101

chúc bn 

hc tốt

\(\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{22}{45}.x=\frac{22}{45}\Rightarrow x=1\)

bạn tách ra xong làm cx dễ mà đây là toán 6

Cảm ơn câu trả lời thật súc tích và thật ngắn gọn của bạn

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(A=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)

~ Hok tốt ~

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\cdot\frac{22}{45}=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}=1\)

\(\Rightarrow x=2\)

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{512}\)

\(=\frac{255}{512}\)

Vậy \(A=\frac{255}{512}\)

A=14 +18 +116 +132 +164 +1128 +1256 +1512 

=12 −14 +14 −18 +....+1256 −1512 

=12 −1512 

=255512 

Vậy A=255512 

Phạm Long Khánh

A = 255/512

TẦM NHƯ HƠI CĂNG

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)

\(=\frac{1}{2000}\)

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

<=>  \(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)

<=>  \(94< x< 92\)vô lí

Vậy không tìm đc x thỏa mãn

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(=\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow94< x< 92\)

\(\Rightarrow\)ĐỀ SAI

Đề bài sai .