(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023

\(a,50\%+\dfrac{7}{12}-\dfrac{1}{2}\\ =\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\\ =\dfrac{7}{12}\\ b,2022\times67+2022\times43-2022\times10\\ =2022\times\left(67+43-10\right)\\ =2022\times100\\ =202200.\\ c,125-25:3\times12\)

\(=25\times5-25:3\times12\\ =25\times\left(5-\dfrac{1}{3}\right)\times12\\ =25\times\dfrac{14}{3}\times12\\ =1400\)

a,50%+127​−21​=21​+127​−21​=(21​−21​)+127​=127​b,2022×67+2022×43−2022×10=2022×(67+43−10)=2022×100=202200.c,125−25:3×12

=25×5−25:3×12=25×(5−13)×12=25×143×12=1400=25×5−25:3×12=25×(5−31​)×12=25×314​×12=1400

a) 7/12

b) 202200

c) 1400

d) mình ko bt

( 1 + 1/2 ) * ( 1 + 1/3 ) * ... * ( 1 + 1/2022 )

= 3/2 * 4/3 * ... * 2023/2022

= 2023/2

1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)

\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)

\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)

\(=5-\dfrac{4}{2}\)

\(=5-2\)

\(=3\)

b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)

\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)

\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)

\(=0+0+0+2022\)

\(=2022\)

2) \(0,7^2\cdot x=0,49^2\)

\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)

\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)

\(\Rightarrow x=\left(0,7\right)^2\)

\(\Rightarrow x=0,49\)

b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)

\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)

\(\Rightarrow x=\left(-0,5\right)^5\)

\(\Rightarrow x=-\dfrac{1}{32}\)

2:

a: =>x*0,49=0,49^2

=>x=0,49

b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5

ai biết làm ko

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2021}{2022}\\ \dfrac{1}{2022}\)

câu a chưa đủ đề em hấy

c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0

       (\(x\) - 2022).(\(x\) - 4) = 0

         \(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)

b, (\(x\) - 1)(\(x^2\) + \(x\) + 1) - \(x\)(\(x\) - 2)(\(x\) + 2)  = 7

     \(x^3\) - 1 - \(x\).(\(x^2\) - 4) = 7

      \(x^3\) - 1 - \(x^3\) + 4\(x\)  = 7

        (\(x^3\) - \(x^3\)) - 1 + 4\(x\) = 7

                       - 1 + 4\(x\) = 7

                               4\(x\) = 7 + 1

                               4\(x\) = 8

                                  \(x\) = 8:4

                                   \(x\) = 2

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2023}\)
\(\Rightarrow x+1=2023\)
\(\Rightarrow x=2022\)
Vậy x = 2022
#kễnh

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}\)

= \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{x+1-x}{x.\left(x+1\right)}\)

= \(\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+...+\dfrac{x+1}{x.\left(x+1\right)}-\dfrac{x}{x.\left(x+1\right)}\)

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

= \(1-\dfrac{1}{x+1}\) =\(\dfrac{2022}{2023}\)

= \(\dfrac{2023}{2023}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)

⇒ \(x+1=2023\)

\(x=2023-1=2022\)