khổ thân cậu bé.bé thế mà khổ

sory về hoàn cảnh của bạn. Hu Hu. Tớ giận mình đã ko giúp được bạn.

mik chỉ giải được bài 1 với bài 2 thôi!!! 

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

\(A=\frac{2^{100}-1}{2^{100}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

hok tốt!!

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Vậy \(A=1-\frac{1}{2^{100}}\)

Đáp án đúng là : \(\frac{9}{196}\) nha bạn

đáp án là:\(\frac{9}{196}\)

Đặt A = \(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\)

5A = \(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\)

5A = \(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\)

5A = \(\frac{1}{4}-\frac{1}{49}\)

5A = \(\frac{45}{196}\)

A = \(\frac{45}{196}:5\)

A = \(\frac{9}{196}\)

mình không biết nữa bằng bao nhiêu ấy nhỉ .......? .......? Sory ^.^

1/3 + 13/15 + 33/35 + 61/63 + 97/99

= 45/11 ( mình không tiện giải, để khi khác giải sau)

Chúc bạn may mắn!

= 45/11

mik làm biếng ghi lâu lắm bạn ạ !!!

k mik nhaaaaaaaaaaaaaaaaaaaaaa

1.A= 1.2.3+2.3.4+...+29.30.31+x=15

\(4A=1.2.3.4+2.3.4.\left(5-1\right)+...+29.30.31.\left(32-28\right)+4x=60\)

\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+29.30.31.32-28.29.30.31+4x=60\)

Từ đó suy ra nha bạn

2.\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(=\frac{2}{2\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(=2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\\ =1-\frac{2}{\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{2009}\Rightarrow x+1=2009\Rightarrow x=2008\)

D = $\frac{2}{3}.\frac{5}{6}.\frac{9}{10}. ... .\frac{799}{780}$

   = $\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}. ... .\frac{38.41}{39.40}$

   = $\frac{2.2}{2.3}.\frac{2.3. ... .38}{3.4. ... 39}.\frac{5.6. ... .41}{4.5. ... .40}$

   = $\frac{2}{3}.\frac{2}{39}.\frac{41}{4}$

   = $\frac{41}{3.39}$

D = \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}.....\frac{779}{780}\)

   = \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}.....\frac{38.41}{39.40}\)

   = \(\frac{2}{3}.\frac{2.3.4....38}{3.4.5....39}.\frac{5.6.7.....41}{4.5.6.....40}\)

   = \(\frac{2}{3}.\frac{2}{39}.\frac{41}{4}\)

   = \(\frac{41}{117}\)

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{9999}\)

=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)

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

=\(1-\frac{1}{101}=\frac{100}{101}\)

tích trước đi đã!!!!!!!!!

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)

\(2\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)

\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(2\left(1-\frac{1}{101}\right)\)

\(2\times\frac{100}{101}\)

\(\frac{200}{101}\)