Tìm x biết:
\(\frac{x+2}{2021}\) + \(\frac{x+5}{2024}\) + \(\frac{x+3}{2022}\) = 3
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BC
5
NH
16 tháng 7 2019
\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2020}\)
\(\Leftrightarrow(\frac{x+4}{2019}+1)+(\frac{x+3}{2020}+1)=(\frac{x+2}{2021}+1)+(\frac{x+1}{2022}+1)\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{x+2023}{2022}\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)
\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2023=0\)
\(\Leftrightarrow x=-2023\)
ND
2
21 tháng 8 2019
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\Rightarrow-\frac{1}{12}\left(x-2018\right)=0\Leftrightarrow x=2018\)
B
4 tháng 9 2020
Bài làm :
Ta có :
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\text{Vì : }\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\ne0\Rightarrow x-2018=0\)
\(\Rightarrow x=2018\)
Vậy x=2018
2
BM
25 tháng 8 2019
ko ghi lại đề
ta thấy : 2019 - 1 = 2018
2020 - 2 = 2018
2021 - 3 = 2018
2022 - 4 = 2018
=> x = 2018
thử lại :
2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022
= 1 + 1 = 1 + 1
2 = 2
11 tháng 4 2023
1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
| 2x+1 | 1 | 3 | 7 | 21 |
| x | 0 | 1 | 3 | 10 |
| TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
2 tháng 4 2018
Ta có :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+\frac{x-4}{2009}+\frac{x-2021}{2}=0\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+\left(\frac{x-4}{2009}-1\right)+\left(\frac{x-2021}{2}+4\right)=0\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+\frac{x-2013}{2009}+\frac{x-2013}{2}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\ne0\)
Nên \(x-2013=0\)
\(\Rightarrow\)\(x=2013\)
Vậy \(x=2013\)
Chúc bạn học tốt ~
3 tháng 1 2018
\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)
\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)
Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
3 tháng 1 2018
=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)
=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018
=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0
=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0
=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )
=> x=-2017
Vậy x=-2017
k mk nha
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 4
Bài 1:
A = 1996 x 1997 x 1998 x 1999 + 2021 x 2022 x 2023 x 2024
A = (1996 x 1997) x (1998 x 1999) + (2021 x 2022) x (2023 x 2024)
A = \(\overline{..2}\) x \(\overline{..2}\) + \(\overline{..2}\) x \(\overline{..2}\)
A = \(\overline{..4}\) + \(\overline{..4}\)
A = \(\overline{..8}\)
Ta có : \(\frac{x+2}{2021}+\frac{x+5}{2024}+\frac{x+3}{2022}=3\)
=> \(\left(\frac{x+2}{2021}-1\right)+\left(\frac{x+5}{2024}-1\right)+\left(\frac{x+3}{2022}-1\right)=3-1-1-1\)
\(\Rightarrow\frac{x-2019}{2021}+\frac{x-2019}{2024}+\frac{x-2019}{2022}=0\)
\(\Rightarrow\left(x-2019\right)\left(\frac{1}{2021}+\frac{1}{2024}+\frac{1}{2022}\right)=0\)
Vì \(\frac{1}{2021}+\frac{1}{2024}+\frac{1}{2022}\ne0\)
=> x - 2019 = 0
=> x = 2019