Đây là bài toán tìm tổng dãy số có quy luật.

Để ý thấy rằng \(\frac{1}{n\left(n+2\right)}=\frac{1}{2}.\frac{2}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)

Vậy thì \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{n+2}\right)=\frac{5}{36}\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)

\(\Rightarrow\frac{1}{n+2}=\frac{1}{18}\Rightarrow n=16.\)

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{36}\)

\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{36}\)

\(\frac{12}{36}-\frac{1}{n+2}=\frac{5}{36}\)

\(\frac{1}{n+2}=\frac{7}{36}\)

\(\Rightarrow\frac{7}{7\left(n+2\right)}=\frac{7}{36}\)

\(7\left(n+2\right)=36\)

n + 2 = 36/7

n = 36/7 - 2

( Tự tính KQ nha )

1/3.5+1/5.7+1/7.9+....+1/n(n+2)=5/36

=1(1/3.5+1/5.7+1/7.9+....+1/n(n+2)

=1/2(2/3.5+2/5.7+2/7.9+....+2/n(n+2)

=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/n-1/n+2

=1/2(1/3+-1/5+1/5+-1/7+1/7+-1/9+...+1/n+-1/n+2

=1/2(1/3+0+0......+-1/n+2)=5/36

=1/2(1/3+1/n+2)=5/36

=(1/3+1/n+2=5/36:1/2

=1/3+1/n+2=5/18

=1/n+2=5/18-1/3

1/n+2=-1/18

=1/n+2=1/-18

=1/n=1/-18-2

=1/n=1/-20

n=-20

=

11.A

12.B

Bài 2 : 

-1/2= -1.30/2.30=-30/60. 

2/3=2.20/3.20=40/60.

-2/5=-2.12/5.12= -24/60.

-3/4=-3.15/4.15=-45/60.

5/4=5.15/4.15=75/60.

Vì 75/60>40/60>-24/60>-30/60>-45/60. 

=>5/4 > 2/3>-2/5>-1/2>-3/4. 

V...

a, -1+3 - 5 + 7 - ...... +97 - 99

[ - 1+ 3] - [ 5 + 7] -  .... - [ 95 + 97] - 99

  [2 - 12] - ..... - [184 - 192] - 99

còn lại tự giải

nnnnnnnnnnnnnnnnnnnnnnnoooooooooooooooooo.......

b

-100

c

-50 

chức bạn học tốt

Kẻ đường trung trực AH của tam giác cân ABC

\(\Rightarrow\)\(\left\{{}\begin{matrix}BH=\dfrac{1}{2}BC\\\widehat{BAH}=\widehat{HAC}=\widehat{BAC}:2=60^0\end{matrix}\right.\)

Ta có: \(AB=AC=6cm\)

Áp dụng tỉ số lượng giác trong tam giác ABH vuông tại H:

\(sinBAH=\dfrac{BH}{AB}\Rightarrow sin60^0=\dfrac{BH}{6}\Rightarrow BH=3\sqrt{3}cm\)

\(\Rightarrow BC=2BH=6\sqrt{3}\left(cm\right)\)

Ta có: \(\widehat{ABC}=\widehat{ACB}=\dfrac{180^0-\widehat{BAC}}{2}=30^0\)(Tam giác ABC cân tại A)

giúp mình vs mn ơi

=\(\dfrac{4}{3}+\dfrac{9}{8}+\dfrac{16}{15}+...+\dfrac{100}{99}\)

=\(\frac{2.2}{1.3}\)+\(\frac{3.3}{2.4}\)+.....+\(\frac{10.10}{9.11}\)

=\(\frac{2.3......10}{1.2.3.....9}\)+\(\frac{2.3.4....10}{3.4.5......11}\)

=\(10+\dfrac{2}{11}\)

=\(10\frac{2}{11}\)