2A =2+\(\frac{3}{2^2}\)+\(\frac{4}{2^3}\)+\(\frac{5}{2^4}\)+.....+\(\frac{100}{2^{99}}\)

\(\Rightarrow\)A=2A-A=1+\(\frac{3}{4}\)\(\frac{1}{2^3}\)+\(\frac{1}{2^4}\)+.....+\(\frac{1}{2^{99}}\)-\(\frac{100}{2^{100}}\)

\(\Rightarrow\)2A=2+\(\frac{3}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+....+\(\frac{1}{2^{98}}\)-\(\frac{100}{2^{99}}\)

\(\Rightarrow\)A=2A-A=1+\(\frac{3}{4}\)+\(\frac{1}{4}\)-\(\frac{101}{2^{99}}\)+\(\frac{100}{2^{100}}\)=2-\(\frac{51}{2^{99}}\)

Đặt biểu thức trong ngoặc đơn là B

\(5B=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}+\dfrac{1}{5^{99}}\)

\(\Rightarrow4B=5B-B=1-\dfrac{1}{5^{100}}\Rightarrow B=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)

\(\Rightarrow A=4.5^{100}.\dfrac{1}{4}\left(\dfrac{5^{100}-1}{5^{100}}\right)+1=\)

\(=5^{100}\)

\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\)

=> \(2A-A=A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Đặt \(B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}\)

=> \(2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{98}}\)

=> \(B=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)

=> \(A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{100}{2^{100}}-\dfrac{1}{2^{99}}\)

=> \(A=2-\dfrac{102}{2^{100}}< 2\)

a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

Bài 3

a,26/100+0,009+41/100+0,24

0,26+0,09+0,41+0,24

(0,26+0,24)+(0,09+0,41)

0,5+0,5

=1

b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4

(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)

30+1+1+1+1

=34

Bài 4,5 khó quá mik ko bít lamf^^))

Bài 5: vì \(\dfrac{3}{11}\) = \(\dfrac{3\times5}{11\times5}\) = \(\dfrac{15}{55}\)

Vậy Khi giữ nguyên tử số thì số cần thêm vào mẫu số là: 

              55 - 39 = 16

Đáp số: 16 

Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1

           \(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)

         b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)

Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)

\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^{100}}\)

\(\dfrac{1}{2}\cdot A=\dfrac{1}{2}+\dfrac{3}{2^4}+...+\dfrac{100}{2^{101}}\)

\(A-\dfrac{A}{2}=\dfrac{1}{2A}=\dfrac{1}{2}+\dfrac{3}{2^3}+...+\dfrac{100}{2^{101}}\)

\(\left[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right]-\dfrac{100}{2^{101}}\) (do 3/2^3=1/2^2+1/2^3)

\(\left[1-\left(\dfrac{1}{2}\right)^{101}\right]\left(1-\dfrac{1}{2}\right)-\dfrac{100}{2^{101}}\)

\(\left(\dfrac{2^{101-1}}{2^{100}}\right)-\dfrac{100}{2^{101}}\)

\(\Rightarrow A=\dfrac{\dfrac{\left(2^{101-1}\right)}{2^{99}-100}}{2^{100}}\)

A=1+B

B=\(\Sigma\left(\dfrac{x}{2^x}\right)\)( cho x chạy từ 3 đến 100) =1

=> A=1+B=1+1=2

\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+.......+\dfrac{100}{2^{100}}\)

\(\Leftrightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+.........+\dfrac{100}{2^{99}}\)

\(\Leftrightarrow2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+......+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+.........+\dfrac{100}{2^{100}}\right)\)

\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Đặt :

\(H=\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}\)\(\Leftrightarrow A=\dfrac{11}{4}-H-\dfrac{100}{2^{100}}\)

\(\Leftrightarrow2H=\dfrac{1}{2^2}+\dfrac{1}{2^3}+........+\dfrac{1}{2^{98}}\)

\(\Leftrightarrow2H-H=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{99}}\right)\)

\(\Leftrightarrow H=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)

\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^2}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Lời giải:

Gọi phân số vế trái là $A$. Gọi tử số là $T$. Xét mẫu số:
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+....+1-\frac{1}{100}\)

\(=99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=100-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})\)

\(=\frac{1}{2}\left[200-(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100})\right]=\frac{1}{2}T\)

$\Rightarrow A=\frac{T}{\frac{1}{2}T}=2$ 

Ta có đpcm.

Giải:

Vì \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\) nên phần tử gấp 2 lần phần mẫu

Ta có:

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)  

\(=\dfrac{2.\left[100-\left(\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)

\(=\dfrac{2.\left[\left(2-\dfrac{3}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{5}\right)+...+\left(1-\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=\dfrac{2.\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+...+\dfrac{99}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=2\) 

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\left(đpcm\right)\) 

Chúc bạn học tốt!