Lời giải:

Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+....+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)

\(7^2A=1-\frac{1}{7^2}+....+\frac{1}{7^{4n-4}}-\frac{1}{7^{4n-2}}+...+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)

\(\Rightarrow A+7^2A=1-\frac{1}{7^{100}}\Rightarrow 50A=1-\frac{1}{7^{100}}<1\)

$\Rightarrow A< \frac{1}{50}$

Đặt \(S=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\)

\(\Rightarrow\dfrac{S}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\)

\(\Rightarrow S+\dfrac{S}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\right)\)

\(\Leftrightarrow\dfrac{50S}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}< \dfrac{1}{50}\)

\(\Leftrightarrow S< \dfrac{1}{50}\)

Vậy \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}< \dfrac{1}{50}\) (Đpcm)

Đặt \(A=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\)

Ta có:

\(\dfrac{A}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\)

\(\Rightarrow A+\dfrac{A}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\right)\)

\(\Rightarrow\dfrac{50A}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}\)

\(\Rightarrow A< \dfrac{1}{50}\)

=> ĐPCM.

\(\text{Đặt:}S=\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\Rightarrow49S=1-\frac{1}{7^2}+.....-\frac{1}{7^{98}}\Rightarrow49S+S=50S=\left(1-\frac{1}{7^2}+\frac{1}{7^4}-....-\frac{1}{7^{98}}\right)+\left(\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\right)=1-\frac{1}{7^{100}}< 1\Rightarrow S< \frac{1}{50}\left(\text{đpcm}\right)\)

Đặt \(A=\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}\)

\(7A=\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{99}}\)

\(\Rightarrow7A-A=\dfrac{1}{7}-\dfrac{1}{7^{100}}\)

\(\Rightarrow6A=\dfrac{1}{7}-\dfrac{1}{7^{100}}\)

\(\Rightarrow A=\dfrac{1}{6}\left(\dfrac{1}{7}-\dfrac{1}{7^{100}}\right)\)

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)

\(=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

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 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)

Mình làm được một câu thôi, bạn dựa vào làm nha!