a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)

b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)

\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)

\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)

\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)

a) (x-2)(x+3) <0 => x-2 và x+3 phải trái dấu

=> x-2<0 và x+3>0

hoặc x-2>0 và x+3<0

=> x<2 và x>-3 => -3<x<2

hoặc x>2 và x<-3 ( vô lý ) ( loại )

=> x \(\in\) { -2;-1;0;1 }

Đúng 100%, tích nha, please!!

\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)

Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{9.11}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(A=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

Thay A vào biểu thức

\(\Rightarrow\frac{5}{11}x=\frac{2}{3}\)

\(\Rightarrow x=\frac{22}{15}\)

P/s: Có thể tính sai :(

\(\left[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right]\times x=\frac{2}{3}\)

Trước tiên mình tính dãy có dấu ngoặc đã

Đặt : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

\(=\frac{1}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right]\)

\(=\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right]\)

\(=\frac{1}{2}\left[1-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{11}\right]\)

\(=\frac{1}{2}\left[1-\frac{1}{11}\right]=\frac{1}{2}\cdot\frac{10}{11}=\frac{1\cdot10}{2\cdot11}=\frac{1\cdot5}{1\cdot11}=\frac{5}{11}\)

Thay vào biểu thức \(S=\frac{5}{11}\)ta lại có :

\(\frac{5}{11}\times x=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{2}{3}:\frac{5}{11}\)

\(\Leftrightarrow x=\frac{2}{3}\cdot\frac{11}{5}\)

\(\Leftrightarrow x=\frac{22}{15}\)

Vậy \(x=\frac{22}{15}\)

\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)

\(\Rightarrow3x-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)=0\)

\(\Rightarrow3x-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)=0\)

\(\Rightarrow3x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)

\(\Rightarrow3x-\left(1-\frac{1}{99}\right)=0\)

\(\Rightarrow3x-\frac{98}{99}=0\)

\(\Rightarrow3x=0+\frac{98}{99}\)

\(\Rightarrow3x=\frac{98}{99}\)

\(\Rightarrow x=\frac{98}{99}:3\)

\(\Rightarrow x=\frac{98}{297}\)

\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)

\(2\left(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}\right)=2.0\)

\(6x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}=0\)

\(6x-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=0\)

\(6x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)

\(6x-\left(1-\frac{1}{11}\right)=0\)

\(6x-\frac{10}{11}=0\)

\(6x=\frac{10}{11}\)

\(x=\frac{5}{33}\)

Tk mình đi mọi người mình bị âm nè!

ai tk mình mình tk lại cho!!!

( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 ) x X = 2/3

                                                       X = 2/3 : ( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 )

                                                       X = 286/85

k mình đi mình đang bị âm

x=1,466666667

<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)

<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)

<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)

<=> \(\frac{2}{15}+x=\frac{17}{15}\)

=> x = 1

(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15

[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15

[2.(1/3-1/15)]+x=17/15

(2.4/15)+x=17/15

6/15+x=17/15

x=17/15-6/15

x=11/15

nhìn công thức đây này \(\sqrt[]{\sqrt{ }\hept{\begin{cases}\\\end{cases}}\frac{ }{ }^{ }_{ }\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}}\) xong rồi đó không cần cảm ơn

(1/15+1/35+1/63+1/99).x=4

4/33.x=4

x=4:4/33

x=16/33

Ta có \(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).x=4\)

=> \(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=4\)

=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=4\)

=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).x=8\)

=> \(\left(\frac{1}{3}-\frac{1}{11}\right).x=8\)

=> \(\frac{8}{33}.x=8\)

=> x = 33

Vậy x = 33

\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)

\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

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

\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)

\(\frac{10}{11}.y=\frac{4}{3}\)

\(\Rightarrow y=\frac{22}{15}\)