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x + (x + 1) + (x + 2) + ... + (x + 2022) + 2022 = 2022
x + x + x + ... + x + 1 + 2 + 3 + ... + 2022 + 2022 = 2022 (1)
Số số hạng x:
2022 - 0 + 1 = 2023 (số)
Từ (1) ta có:
2023x + 2022.2023 : 2 + 2022 = 2022
2023x + 2045253 = 2022 - 2022
2023x = 0 - 2045253
2023x = -2045253
x = -2045253 : 2023
x = -1011
Ta có : x + (x + 1) + (x + 2) + ... + (x+2022) + 2022 = 2022
=> x + (x + 1) + (x + 2) + ... + (x + 2022) = 2022 - 2022
=> [x + (x + 2022) ] . { [ (x + 2022) - x) : 1 + 1] } : 2 = 0
( số đầu + số cuối . số số hạng : 2 )
=> (2x + 2022) . 2023 : 2 = 0
=> 2x + 2022 = 0 . 2 : 2023= 0
=> (2x + 2022) : 2 = 0 : 2
=> x + 1011 = 0 => x = -1011
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2023}\)
\(\Rightarrow x+1=2023\)
\(\Rightarrow x=2022\)
Vậy x = 2022
#kễnh
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}\)
= \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{x+1-x}{x.\left(x+1\right)}\)
= \(\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+...+\dfrac{x+1}{x.\left(x+1\right)}-\dfrac{x}{x.\left(x+1\right)}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
= \(1-\dfrac{1}{x+1}\) =\(\dfrac{2022}{2023}\)
= \(\dfrac{2023}{2023}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
⇒ \(x+1=2023\)
\(x=2023-1=2022\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). 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
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{505}{1011}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1010}{1011}\)
=>1/x+1=-1009/2022
=>x+1=-2022/1009
hay x=-3031/1009
a, ( 13.x - 122) : 5 = 5
( 13.x - 122) = 5.5
( 13.x - 122) = 25
( 13.x - 144) = 25
13.x = 25 + 144
13.x = 169
x = 169 : 13
x = 13
Vậy x = 13
b, 3.x[82 - 2.(25 - 1)] = 2022
3.x[64 - 2.(32 - 1)] = 2022
3.x[62 - 2.31] = 2022
3.x[62 - 62] = 2022
3.x.0 = 2022
3.x = 2022 : 0
3.x = 0
x = 0 : 3
x = 0
Vậy x = 0
Đây bạn nhé !!!
Chúc bạn học tốt !!!
\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)
\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
(x-1)2020=(x-1)2022
=>(x-1)2020-(x-1)2022=0
=>(x-1)2020-(x-1)2020.(x-1)2=0
=>(x-1)2020(1-(x-1)2=0
=>(x-1)2020=0 hoặc 1-(x-1)2=0
=>x=1 hoặc x=2.
Bài 2
a,2105 và 545
2105=(27)15=12815
545=(53)15=12515
Vì 12815>12515 nên 2105>545.
b,
554 và 381
554=(56)9=156259
381=(39)9=196839
Vì 156259<196839 nên 554<381
Bài 1 :
\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)
\(\Rightarrow\left(x-1\right)^{2022}-\left(x-1\right)^{2020}=0\)
\(\Rightarrow\left(x-1\right)^{2020}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
1/2×X+1/3×X+X=2022
= 1/2×X+1/3×X+Xx1 =2022
= X x ( 1/2 + 1/3 + 1) = 2022
= X x 11/6 = 2022
= X = 2022: 11/6
= X = 12132/11
\(\dfrac{1}{2}\times x+\dfrac{1}{3}\times x+x=2022\\ \Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+1\right)\times x=2022\\\Rightarrow \left(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{6}{6}\right)\times x=2022\\ \Rightarrow\dfrac{11}{6}\times x=2022\\ \Rightarrow x=2022:\dfrac{11}{6}\\ \Rightarrow x=2022\times\dfrac{6}{11}\\ \Rightarrow x=\dfrac{12132}{11}\)