\(S=\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{998.1000}\right)=\)

\(=998+\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{997.999}\right)+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{998.1000}\right)\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{997.999}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{997.999}=1-\dfrac{1}{999}=\dfrac{998}{999}\)

\(\Rightarrow A=\dfrac{488}{999}\)

\(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{998.1000}\)

\(\Rightarrow2B=\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{998.1000}=\dfrac{1}{2}-\dfrac{1}{1000}=\dfrac{499}{1000}\)

\(\Rightarrow B=\dfrac{499}{2000}\)

\(\Rightarrow S=998+\dfrac{499}{999}+\dfrac{499}{2000}\)

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

=>OA+AB=OB

=>AB+3=6

=>AB=3(cm)

b: A nằm giữa O và B

mà AO=AB(=3cm)

nên A là trung điểm của OB

vế kia đâu

1/21 + 1/28 + 1/36 + .... + 2/x.(x+1) = ?

?=2/9

bạn trả lời câu b bao gồm cỡ giày 38,39,40,41 có đúng ko?

a: Tổng số giày bán được trong tháng 5 là:

\(18+32+58+65+42+15=230\left(cái\right)\)

b: Cửa hàng nên nhập về nhiều hơn những cỡ giày 39;40;41

\(\dfrac{1}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

=>\(\dfrac{3+xy}{3x}=\dfrac{5}{6}\)

=>\(6\left(xy+3\right)=5\cdot3x\)

=>\(2\left(xy+3\right)=5x\)

=>2xy-5x=-6

=>x(2y-5)=-6

mà 2y-5 lẻ

nên \(\left(x;2y-5\right)\in\left\{\left(6;-1\right);\left(-6;1\right);\left(2;-3\right);\left(-2;3\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(6;2\right);\left(-6;3\right);\left(2;1\right);\left(-2;4\right)\right\}\)

\(\left(-4,44+60-5,56\right):\left(-2\right)\)

\(=\left(60-10\right):\left(-2\right)\)

\(=\dfrac{50}{-2}=-25\)

\(A=\dfrac{3}{2^2}+\dfrac{8}{3^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(=1+1+...+1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}\right)\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{2023^2}< \dfrac{1}{2022\cdot2023}=\dfrac{1}{2022}-\dfrac{1}{2023}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1\)

=>\(0< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1\)

=>A không là số tự nhiên

A=3/2^2 + 8/3^2 + ... + 2023^2 - 1/2023^2

A =2^2-1/2^2  + 3^2-1/3^2 +...+ 2023^2-1/2023^2

A=1 - 1/2^2 + 1- 1/3^2 + ... + 1 - 1/2023^2

A=1+1+...+1 - (1/2^2 +1/3^2 + 1/4^2 +...+1/2023^2)

A=2022 - (1/2^2 + 1/3^2 + ... + 1/2023^2) <2022 (1)

Ta có 1/2^2 < 1/1.2

           1/3^2 <1/2.3

           .................

            1/2023^2 < 1/2022.2023

suy ra 

1/2^2 + 1/3^2 + ... +1/2023^2 <1/1.2 + 1/2.3 +...+1/2022.2023

Ta có 

1/1.2 + 1/2.3 + .... +1/2022.2023

=1/1 - 1/2 + 1/2 - 1/3 + ....+1/2022 - 1/2023

=1/1 - 1/2023

suy ra 1/2^2 + 1/3^2 + ... + 1/2023^2<1-1/2023

suy ra A =2022 - (1/2^2 + 1/3^2 + .... + 1/2023^2) > 2022-(1-2023)

suy ra 2022 - (1/2^2 + 1/3^2 +...+1/2023^2) >2021 + 1/2023 >2021(2)

tù 1,2 suy ra 

    2021<A<2022

 suy ra A ko là số tự nhiên 

Vậy A ko là số tự nhiên

\(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}\)

\(=\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}\)

\(=2\left(\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}\right)\)

\(=2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)\)

\(=2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{1}{3}-\dfrac{1}{2\left(x+1\right)}\)

Thời gian học toán trong 1 ngày của Hà là:

\(\dfrac{5}{24}\times\dfrac{1}{2}=\dfrac{5}{48}\)(ngày)