1.

a) \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)

\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)

\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{6}{2}.\frac{10}{39}\)

\(=\frac{10}{13}\)

b) \(\frac{3}{24}+\frac{3}{48}+\frac{3}{80}+\frac{3}{120}+\frac{3}{168}\)

\(=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)

\(=\frac{3}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{14}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)

\(=\frac{3}{2}.\frac{5}{28}\)

\(=\frac{15}{56}\)

\(a.\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{11.13}\)

\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

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

\(=3.\frac{10}{39}\)

\(=\frac{10}{13}\)

\(a.\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)

\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)

\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

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

\(=3.\frac{10}{39}\)

\(=\frac{10}{13}\)

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

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

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

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

1) `-3\sqrt13=-3\sqrt13`

`-9=-3\sqrt9`

`\sqrt13>\sqrt9`

`=> -3\sqrt13 < -3\sqrt9`

`=> -3\sqrt13 < 9`.

2) `\sqrt15 < \sqrt16`

`<=> \sqrt15-1 < \sqrt16-1`

`<=> \sqrt15-1 < 3 < \sqrt10`

`=> \sqrt15-1 <\sqrt10`

3) `5=4+1=\sqrt16+1`

`\sqrt8+1=\sqrt8+1`

`=> 5>\sqrt8+1`

1) \(-3\sqrt{13}=-\sqrt{117}< -\sqrt{81}=-9\)

3) Ta có: \(5^2=25=9+16\)

\(\left(2\sqrt{2}+1\right)^2=9+4\sqrt{2}\)

mà \(16>4\sqrt{2}\)

nên \(5>2\sqrt{2}+1\)

1) $-3\sqrt{13}=-\sqrt{117}<-\sqrt{81}=-9$

3) Ta có: ${}^{}=25=9+16$

${}^{}=9+4\sqrt{2}$

mà $16>4\sqrt{2}$

nên $5>2\sqrt{2}+1$

A=1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15

A=1/1 - 1/3 +1/3 - 1/5 +1/5 -1/7+......+1/13 - 1/15

A=1 - 1/15

A=1/14

bạn ơi: 

1/195 mới bằng 1/13.15 mà!!

1/3+1/15+1/35+1/63+1/99+1/143+1/180=1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

=1/2.(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+.....+1/13-1/15)

=1/2.(1-1/15)

=1/2.14/15

=7/15

chúc học tốt

Giải: 
Đặt A = 1/3+1/15+1/35+1/63+1/99+1/143+1/195 
2A= 2/(1.3) + 2/(3.5) + 2/(5.7) + 2/(7.9)+2/(9.11) + 2/(11.13)+2/(13.15) 
2A=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9... 
2A=1/1-1/15=14/15 
Vậy A=14/15 : 2 = 7/15

Nhấn đúng mk nha             Tran Dan 

  \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+..+\frac{1}{143}+\frac{1}{195}\)

=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{13.15}\)

= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{13}-\frac{1}{15}\)

= \(1-\frac{1}{15}=\frac{14}{15}\)

tick đúng nha 

A = \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{13\times15}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}\right)\)

A = \(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\times\left(1-\frac{1}{15}\right)=\frac{7}{15}\)

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

BÀI 2

60%.x + 0,4.x + x:3= 2

\(\frac{60}{100}\)x + \(\frac{4}{10}\).x + x. \(\frac{1}{3}\)=2

\(\frac{3}{5}\).x + \(\frac{2}{5}\).x + x.\(\frac{1}{3}\)=2 

(\(\frac{3}{5}\)+ \(\frac{2}{5}\)+ \(\frac{1}{3}\)) .x =2

\(\frac{4}{3}\).x                          =2

        x                            = 2: \(\frac{4}{3}\)

          x                          = \(\frac{3}{2}\)

Vậy x=\(\frac{3}{2}\)

k cho mik nha các bạn

Bài 2 :

60%x + 0.4x + x : 3 = 2

\(x.\left(\frac{60}{100}+\frac{2}{5}+\frac{1}{3}\right)\)= 2

\(x.\frac{4}{3}\)= 2

\(x=2.\frac{3}{4}\)

\(x=1.5\)

Đặt phép tính cần tìm là A

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(2A=1-\dfrac{1}{13}\)

\(2A=\dfrac{12}{13}\)

\(A=\dfrac{6}{13}\)

\(A=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{143}\\ =\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\\ =\dfrac{1}{2}\times\dfrac{12}{13}\\ =\dfrac{6}{13}\)