Lời giải:
\(299A=\frac{300-1}{1.300}+\frac{301-2}{2.301}+\frac{302-3}{3.302}+....+\frac{400-101}{101.400}\)

\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)

\(=(1+\frac{1}{2}+....+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(1)\)

Mặt khác:

$101B=\frac{102-1}{1.102}+\frac{103-2}{2.103}+...+\frac{400-299}{299.400}$

$=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}$

$=(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{299})-(\frac{1}{102}+\frac{1}{103}+....+\frac{1}{400})$

$=(1+\frac{1}{2}+...+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(2)$

Từ $(1);(2)\Rightarrow 299A=101B$

$\Rightarrow \frac{A}{B}=\frac{101}{299}$

tự xử đi

a)\(\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{3}-1\right):...:\left(\dfrac{1}{50}-1\right)=-\dfrac{1}{2}:\left(-\dfrac{2}{3}\right):\left(-\dfrac{3}{4}\right)...:\left(-\dfrac{49}{50}\right)=-\dfrac{1}{2}\cdot\left(-\dfrac{3}{2}\right)\cdot\left(-\dfrac{4}{3}\right)...\left(-\dfrac{50}{49}\right)=-\dfrac{1\cdot3\cdot4...50}{2\cdot3\cdot...\cdot49}=-\dfrac{50}{2}=-25\)

b)Sai đề bạn xem lại và đăng lại mình giải cho

Giải:

a)1/5.8+1/8.11+...+1/x.(x+1)=101/1540

1/3.(3/5.8+3/8.11+...+3/x.(x+1))=101/1540

1/3.(1/5-1/8+1/8-1/11+...+1/x-1/x+1)=101/1540

                              1/3.(1/5-1/x+1)=101/1540

                                      1/5-1/x+1=101/1540:1/3

                                      1/5-1/x+1=303/1540

                                            1/x+1=1/5-303/1540

                                            1/x+1=1/308

⇒x+1=308

       x=308-1

       x=307

b)1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=2020/2021

1/1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2020/2021

                                    1/1-1/x+1=2020/2021

                                          1/x+1=1/1-2020/2021

                                          1/x+1=1/2021

⇒x+1=2021

      x=2021-1

      x=2020

Mk thấy đề bài hơi sai là:

1/x+(x+1) ➜ 1/x.(x+1)

mới ra đc kết quả!

\(C=\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2-1}\right)...\left(\dfrac{1}{101^2-1}\right)\)

\(C=\dfrac{1-200^2}{200^2}.\dfrac{1-199^2}{199^2}.\dfrac{1-198^2}{198^2}...\dfrac{1-101^2}{101^2}\)

\(C=\dfrac{\left(1-200\right)\left(1+200\right)}{200^2}.\dfrac{\left(1-199\right)\left(1+199\right)}{199^2}...\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}.\dfrac{\left(1-101\right)\left(1+101\right)}{101^2}\) \(C=\dfrac{-199.201}{200.200}.\dfrac{-198.200}{199.199}.\dfrac{-197.199}{198.198}...\dfrac{-99.101}{100.100}.\dfrac{-100.102}{101.101}\)

\(C=\dfrac{199.201}{200.200}.\dfrac{198.200}{199.199}.\dfrac{197.199}{198.198}...\dfrac{99.101}{100.100}.\dfrac{100.102}{101.101}\)

\(\Rightarrow C=\dfrac{200}{2.101}=\dfrac{201}{202}\)

Câu 2 mik chịu r sorry:(

A/B =1

Không chắc vì tính nhẩm

Kết quả là \(\dfrac{101}{299}\). Cô mình chữa rồi đó, và mình lúc đầu cũng không làm được.

Ta có :

\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+.............+\dfrac{1}{101.400}\)

\(299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...................+\dfrac{299}{101.400}\)

\(299A=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+..............+\dfrac{1}{101}-\dfrac{1}{400}\)

\(299A=\left(1+\dfrac{1}{2}+................+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+..............+\dfrac{1}{400}\right)=C\)

\(\Rightarrow A=\dfrac{C}{299}\)

Lại có :

\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+................+\dfrac{1}{299.400}\)

\(101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...............+\dfrac{101}{299.400}\)

\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+..................+\dfrac{1}{299}-\dfrac{1}{400}\)

\(101B=\left(1+\dfrac{1}{2}+..............+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}-\dfrac{1}{103}+...............+\dfrac{1}{400}\right)=C\)

\(\Rightarrow B=\dfrac{C}{101}\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{C}{101}:\dfrac{C}{299}=\dfrac{101}{299}\)

~ Chúc bn học tốt ~

::((

so hard

\(A=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2019}+\left(\dfrac{1}{3}\right)^{2020}\)

\(\Rightarrow\dfrac{1}{3}A=\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2021}\)

\(\Rightarrow\dfrac{2}{3}A=A-\dfrac{1}{3}A=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2020}-\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^{2021}=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^{2021}< \dfrac{1}{3}\)

\(\Rightarrow A< \dfrac{1}{2}\)