\(-\frac{12}{35}\div\frac{7}{11}-\frac{23}{35}\div\frac{7}{11}-\frac{5}{11}\)

\(=\left(-\frac{12}{35}-\frac{23}{35}\right)\div\frac{7}{11}-\frac{5}{11}\)

\(=-1\div\frac{7}{11}-\frac{5}{11}\)

\(=-\frac{11}{7}-\frac{5}{11}\)

\(=-\frac{156}{77}\)

\(2-\frac{11}{9}\div\frac{5}{14}-\frac{7}{9}\times\frac{14}{5}\)

\(=2-\frac{154}{45}-\frac{98}{45}\)

\(=-\frac{64}{45}-\frac{98}{45}\)

\(=-\frac{18}{5}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\)

\(=1-\dfrac{1}{40}\)

\(=\dfrac{39}{40}\)

1/(1×4) + 1/(4×7) + ... + 1/(34×37) + 1/(37×40)

= 1/3 × (1 - 1/4 + 1/4 - 1/7 + ... + 1/34 - 1/37 + 1/37 - 1/40)

= 1/3 × (1 - 1/40)

= 1/3 × 39/40

= 13/40

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)

\(=1-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=1-\dfrac{1}{40}\)

\(=\dfrac{39}{40}\)

Vậy giá trị cần tìm là: \(\dfrac{39}{40}\)

\(3xA=\dfrac{4-1}{1x4}+\dfrac{7-4}{4x7}+...+\dfrac{37-34}{34x37}+\dfrac{40-37}{37x40}=\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{17}{ }+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}=\)

\(=1-\dfrac{1}{40}=\dfrac{39}{40}\Rightarrow A=\dfrac{39}{40}:3=\dfrac{13}{40}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{34.37}+\dfrac{3}{37.40}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=\dfrac{1}{3}.\dfrac{39}{40}\)

\(=\dfrac{13}{40}\)

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

= \(\frac{1}{3}\left(1-\frac{1}{x+1}\right)\)

=\(\frac{1}{3}x\left(\frac{x+1}{x+1}-\frac{1}{x+1}\right)\)

= \(\frac{1}{3}x\frac{x}{x+1}\)

dè bài la tim x

Tong cua phep. Tinh la 2009/6030

Ta có :

1/1.4+1/4.7+...+1/91.94

=1/3.(1/1-1/4+...+1/91-1/94)

=1/3.(1/1-1/94)

=1/3.93/94

=31/94

Ta có :

1/1.4+1/4.7+...+1/91.94

=1/3.(1/1-1/4+...+1/91-1/94)

=1/3.(1/1-1/94)

=1/3.93/94

=31/94

Ta có :

1/1.4+1/4.7+...+1/91.94

=1/3.(1/1-1/4+...+1/91-1/94)

=1/3.(1/1-1/94)

=1/3.93/94

=31/94

mà mình không phải là Thiên Tỷ mình tên là Gia Yến đấy nhé

Giờ anh đang bận hồi nữa anh giúp cho nha

\(F=\dfrac{1}{x}\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}+\dfrac{1}{100.103}\right)\)

\(3F=\dfrac{1}{x}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}+\dfrac{3}{100.103}\right)\)

\(F=\dfrac{\dfrac{1}{x}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)}{3}=\dfrac{\dfrac{1}{x}.\dfrac{100}{309}}{3}=\dfrac{\dfrac{100x}{309}}{3}=\dfrac{100x}{927}\)

\(F=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{204}{309}\)

\(F=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}=\dfrac{34}{103}\)

\(C=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{52.55}+\frac{1}{55.58}\)

\(\Rightarrow C=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{52.55}+\frac{3}{55.58}\right)\)

\(\Rightarrow C=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{52}-\frac{1}{55}+\frac{1}{55}-\frac{1}{58}\right)\)

\(\Rightarrow C=\frac{1}{3}.\left(1-\frac{1}{58}\right)\)

\(\Rightarrow C=\frac{1}{3}.\frac{57}{58}\)

\(\Rightarrow C=\frac{19}{58}\)

Vậy \(C=\frac{19}{58}\)

~ Ủng hộ nhé 

\(3C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{1}{55\times58}..\)

\(3C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}....+\frac{1}{55}-\frac{1}{58}..\)

\(3C=1-\frac{1}{58}=\frac{57}{58}\)

\(C=\frac{57}{174}\)

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