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

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

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

<=> \(\frac{2}{15}+x=\frac{17}{15}\)

=> 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

1/9 xX=1 thì X=9

\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\times x=1\)

\(\left(\frac{63}{945}+\frac{27}{945}+\frac{15}{945}\right)\times x=1\)

\(\left(\frac{63+27+15}{945}\right)\times x=1\)

\(\frac{105}{945}\times x=1\)

\(\frac{1}{9}\times x=1\)

\(x=1:\frac{1}{9}\)

\(x=1\times\frac{9}{1}\)

\(x=9\)

Vậy x=9

=> 2(1/15+1/35+1/63+1/99)x=2

=>(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)x=2

=>8/33x=2

=>x=2:8/33

=>x=8,25

\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\cdot x=1\)

\(\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\cdot x=1\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]\cdot x=1\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\right]\cdot x=1\)

\(\left[\frac{1}{2}\cdot\frac{8}{33}\right]\cdot x=1\)

\(\frac{4}{33}\cdot x=1\)

\(\Rightarrow x=\frac{1}{\frac{4}{33}}=\frac{33}{4}\)

a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)

\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\cdot\frac{8}{33}\)

\(=\frac{52}{33}\)

a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99

A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)

A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)

A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)

A= 13/2 ( 1/3 - 1/11) 

A= 13/2 . 8/33

A= 52/33  

Chuyển vế tất cả số hạng tự do sang phải, ta được \(x=1931\)bạn nhé!

bạn giải cách làm ra luôn nhoa bạn

a)Ta có:
​A= 1/15+1/35+1/63+1/99+1/143
A= 1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
2A= 2/3.5+2/5.7+2/7.9+2/9.11+2/11.13
2A= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
Đơn giản đi ta được:
2A= 1/3-1/13
2A= 10/39
A= 5/39
Vậy A= 5/39   

b) Để A và B có giá trị bằng nhau thì:
\(\frac{3}{4}\cdot x+7=\frac{4}{3}\cdot x-35\)
\(7+35=\frac{4}{3}\cdot x-\frac{3}{4}\cdot x\)
\(42=\frac{7}{12}\cdot x\)
\(x=42:\frac{7}{12}\)
\(x=72\)

= 1/3 x 5 + 1/5x 7 + 1/7 x 9 +...+1/99 x 101

=1/ 2x (1/3 - 1/5 +1/5 - 1/7 +1/7 - 1/9 + 1/99 - 1/101)

=1/2 x (1/3 - 1/99)

=1/2 x (1/3 - 1/101)

=1/2 x (98/303)

=1/15 + 1/35 + 1/63 +1/99+...+1/9999

=49/303 

\(=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)

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

\(=\frac{1}{3}-\frac{1}{101}+0+...+0\)

\(=\frac{98}{303}\)

X = 4.5

Tick nha , chac chan dung

\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\times x=1\)

\(\frac{1}{9}\times x=1\)

\(x=1:\frac{1}{9}=1.9=9\)