\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)

đáp án B bạn nha 

2) \(B=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(1989-1990-1991+1992\right)+1993-1994\)

\(=0+0+...+0+1993-1994=0+1993-1994=-1\)

A=(1-2)+(3-4)+...+(2021-2022)+2023

=2023-(1+1+1+...+1)

=2023-1011

=1012

222222222222222222222222222222222222222222222222222222222222

sos

A=(2x+1-2x+1)^2+xy

=xy+4

=2023+4

=2027

\(A=(2x+1-2x+1)^2+xy\)

\(=xy+4\)

\(=2023+4\)

\(=2027\)

a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145

b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7

c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11

a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21

b) (5/17 x 21/32 x 47/24 x 0)
= 0

c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7

a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x

b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15

c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2

a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36

câu,b,không,đủ,thông,tin,nhan,bạn.

=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023

=0+0+...+0-1-2023

=-2024

Bằng 2024

1. ĐKXĐ: \(x\ne\pm1\)

2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)

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

\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

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

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

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)

$A=\frac{3}{2^2}+\frac{8}{3^2}+\frac{15}{4^2}+....+\frac{{2023}^2-1}{{2023}^2}$

$A=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+....+1-\frac{1}{{2023}^2}$

$A=2022-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2})$

$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2022.2023}$

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2022.2023}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....-\frac{1}{2023}$

$\Rightarrow 0<\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2}<1-\frac{1}{2023}<1$

$\Rightarrow 2022-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2})$ko phải số tự nhiên

$\Rightarrow A$ ko phải số tự nhiên

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k

\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0