mi tích tau tau tích mi xong tau trả lời nka việt nam nói là làm

(tính bằng máy tính)

2/15 + 2/35 + 2/63 + 2/99 + 2/143 + 2/195

\(=2\times\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\right)\)

= \(2\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(=2\times\left(\dfrac{1}{3}-\dfrac{1}{15}\right)\)

\(=2\times\dfrac{4}{15}\)

\(=\dfrac{8}{15}\)

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

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

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

<=> \(\frac{2}{15}+x=\frac{17}{15}\)

=> 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

=2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
= 1/3 - 1/13
= 13/39 -3/39
=10/39 

bằng 10/39 nha

chúc bạn học tốt

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\\ =\dfrac{1}{3}-\dfrac{1}{13}\\ =\dfrac{13}{39}-\dfrac{3}{39}\\ =\dfrac{10}{39}\)

NHỚ TÍNH ĐÚNG VỚI CHO COIN NHA

a)\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{2}{15}\)

b)\(M=1+3+3^2+...+3^{25}=\frac{3^{26}-1}{3-1}=\frac{3^{26}-1}{2}

bạn đọc lại đề bài b) đi

$\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}=\frac{2}{3.5}+\frac{2}{5.7}+$\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}$

=$\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{11}-\frac{1}{13}$

=$\frac{1}{3}-\frac{1}{13}=\frac{10}{39}$

Vậy....

\(S=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)

\(S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(S=\dfrac{1}{3}-\dfrac{1}{13}\)

\(S=\dfrac{13}{39}-\dfrac{3}{39}\)

\(S=\dfrac{10}{39}\)

Vậy \(S=\dfrac{10}{39}\)

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!!