Cậu mới thi viuolympic sao mk cũng mới đăng

đặt \(\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{65.68}\right)\)là A

Ax=\(\frac{19}{68}+\frac{7}{34}=\frac{33}{68}\)

3A=\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{11}{8.11}+...+\frac{1}{65.68}\right)\)

3A=\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\)

3A=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\)

3A=\(\frac{1}{2}-\frac{1}{68}=\frac{33}{68}\)

A=33/68:3=11/68

\(\Rightarrow\)33/68:11/68=3

vậy x= 3

Trả lời giúp mk nha mk cho 3 h lun nói thietj ó

(1/2.5+1/5.8+....+1/65.68)x=19/68+7/34=33/68

(1/2.5+1/5.8+.....+1/65.68).3.x=33/68.3=99/68

(3/2.5+3/5.8+........3/65.68)x =99/68

(1/2-1/68)x=99/68

33/68x=99/68

=>x=3

cho mình nha

bai 1:\(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\right)x-\frac{7}{34}=\frac{19}{68}\)

=\(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)x-\frac{7}{34}=\frac{19}{68}\)

\(\Rightarrow\)x=3

bai2:từ giả thiết \(\frac{ab}{bc}=\frac{a}{c}=\frac{-1}{2}va\frac{ab}{ac}=\frac{b}{c}=\frac{3}{4}\)

hay \(\frac{a}{-2}=\frac{b}{3}=\frac{c}{4}\)

\(\Rightarrow\)\(\left(\frac{a}{-2}\right)^2=\frac{a}{-2}\times\frac{b}{3}=\frac{-6}{-6}=1\)

a=-2 (a<0)

\(\Rightarrow\)a=-2,b=3,c=4

x=1

a= -2

b=3

c=4

ủng hộ mình nha

  \(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)

\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)=\frac{1}{2}\left(\frac{34}{68}-\frac{1}{68}\right)=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)

Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)

\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)

trả lời hộ mik mik cho 1 link

\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+.......+\dfrac{12}{65.68}\)

\(=4\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{65.68}\right)\)

\(=4\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.......+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(=4\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(=4.\dfrac{33}{68}=\dfrac{33}{17}\)

Dễ quá! Vì mình là một CTV bên Học toán với OnlineMath nên bài này easy!! :")))

\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+\dfrac{12}{8.11}+...+\dfrac{12}{65.68}\)

\(\Leftrightarrow4\left(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{65.68}\right)\)

\(\Leftrightarrow4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(=4\left(1-\dfrac{1}{68}\right)=4.\dfrac{67}{68}=\dfrac{67}{17}\)

Sorry giải nhầm bài rồi, tính giải một bài khác mà đi ấn nhầm bài này =((( bạn giải theo ctv kia nhé