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K
Khách
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R
1
S
6 tháng 2 2020
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
NH
6 tháng 2 2020
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
EW
0
VT
1
AH
Akai Haruma
Giáo viên
29 tháng 12 2023
Lời giải:
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$|x-2020|+|x-2024|=|x-2020|+|2024-x|\geq |x-2020+2024-x|=4$
$|x-2022|\geq 0$ (theo tính chất trị tuyệt đối)
$\Rightarrow |x-2020|+|x-2024|+|x-2022|\geq 4+0=4$
$\Rightarrow P\geq 4$
Vậy $P_{\min}=4$. Giá trị này đạt được khi $(x-2020)(2024-x)\geq 0$ và $x-2022=0$
Hay $x=2022$
LG
2
XO
1 tháng 10 2020
Ta có :\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)
=> \(\left(\frac{x+4}{2018}+1\right)+\left(\frac{x+3}{2019}+1\right)=\left(\frac{x+2}{2020}+1\right)+\left(\frac{x+1}{2021}+1\right)\)
=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)
=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)
=> \(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)
Vì \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\)
=> x + 2022 = 0
=> x = -2022
Vậy x = -2022
1 tháng 10 2020
\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)
\(\frac{x+4}{2018}+1+\frac{x+3}{2019}+1=\frac{x+2}{2020}+1+\frac{x+1}{2021}+1\)
\(\frac{x+4}{2018}+\frac{2018}{2018}+\frac{x+3}{2019}+\frac{2019}{2019}=\frac{x+2}{2020}+\frac{2020}{2020}+\frac{x+1}{2021}+\frac{2021}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)
\(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)
\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)
\(x=0-2022\)
\(x=-2022\)
GN
0
DP
19 tháng 7 2017
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\cdot\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{13}\right)\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
d) \(\frac{x+4}{2017}+\frac{x+3}{2018}=\frac{x+2}{2019}+\frac{x+1}{2020}\)
\(\Leftrightarrow\left(\frac{x+4}{2017}+1\right)+\left(\frac{x+3}{2018}+1\right)=\left(\frac{x+2}{2019}+1\right)+\left(\frac{x+1}{2020}\right)\)
\(\Leftrightarrow\frac{x+2021}{2017}+\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
\(\Leftrightarrow\left(x+2021\right)\cdot\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2021=0\)
\(\Leftrightarrow x=-2021\)
19 tháng 7 2017
a) 2/3.X+4=-12
2/3.X=-12-4
2/3.X=-16
x=-16:2/3
x=-16.3/2
x=-24
vậy x=-24
B) 3/4+1/4:X=-3
1/4:X=-3-3/4
1/4:X=-15/4
X=1/4:-15/4
X=1/4.4/-15
X=-1/15
VẬY X=-1/15
MÌNH MỚI NGHĨ RA HAI Ý THÔI.
THÔNG CẢM CHO MÌNH BẠN NHÁ
TL
1
15 tháng 9 2019
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)
=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)
=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)
=> \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)
=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)
Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)
Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025
Ta có :\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2024}{8}+\frac{x-2030}{14}\)
=> \(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2024}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)
=> \(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2016}{8}+\frac{x-2016}{14}\)
=> \(\frac{x-2016}{2}+\frac{x-2016}{4}-\frac{x-2016}{8}-\frac{x-2016}{14}=0\)
=> \(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}-\frac{1}{14}\right)=0\)
=> x - 2016 = 0 (Vì \(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}-\frac{1}{14}\ne0\))
=> x = 2016
Vậy x = 2016
ta có
\(\frac{x-2018}{2}+1+\frac{x-2020}{4}+1=\frac{x-2024}{8}+1+\frac{x-2030}{14}+1\)
\(\Leftrightarrow\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2016}{8}+\frac{x-2016}{14}\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x=2016\)