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HV
18 tháng 1 2020
B=5+2(x-2019)2020
Vì (x-2019)2020 ≥0
=>5+(x-2019)2020 ≥5
Để B đạt Min
=>x-2019=0
=>x=2019
Vậy MinB=5 <=>x=2019
20 tháng 9 2020
+) \(A=3\left(x-4\right)^4-4\ge-4\)
Min A = -4 \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
+) \(B=5+2\left(x-2019\right)^{2020}\ge5\)
Min B = 5 \(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)
+) \(C=5+2018\left(2020-x\right)^2\)
Min C = 5 \(\Leftrightarrow2020-x=0\Leftrightarrow x=2020\)
+) \(D=\left(x-1\right)^{2020}+\left(y+x\right)-1\ge-1\)
Min D = -1 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-x\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)
+) \(E=2\left(x-1\right)^2+3\left(2x-y\right)^4-2\ge-2\)
Min E = -2 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\2x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\2x=y\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
S
2
16 tháng 3 2020
x−42021+x−32020=x−22019+x−12018x−42021+x−32020=x−22019+x−12018
⇔ x−42021+x−32020−x−22019−x−12018=0x−42021+x−32020−x−22019−x−12018=0
⇔ (1+x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0(1+x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0⇔ x+20172021+x+20172020−x+20172019−x+20172018=0x+20172021+x+20172020−x+20172019−x+20172018=0
⇔ (x+2017)(12021+12020−12019−12018)=0(x+2017)(12021+12020−12019−12018)=0
⇔ x + 2017 = 0
⇔ x = -2017
17 tháng 3 2020
\(\frac{x-1}{2020}+\frac{x-2}{2021}=\frac{x+1}{2018}+\frac{x+2}{2017}\)
\(\Leftrightarrow\frac{x-1}{2020}+1+\frac{x-2}{2021}-1=\frac{x+1}{2018}+1+\frac{x+2}{2017}+1\)
\(\Leftrightarrow\frac{x+2019}{2020}+\frac{x+2019}{2021}=\frac{x+2019}{2018}+\frac{x+2019}{2017}\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
mà \(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
S
22 tháng 3 2020
\(x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x\left(x^{2019}-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
S
22 tháng 3 2020
\(1+2+2^2+2^3+....+2^{2019}+2^{2020}\)
\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{2016}+2^{2017}+2^{2018}\right)+2^{2019}+2^{2020}\)
\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+.....+2^{2016}\left(1+2+2^2\right)+2^{2019}+2^{2020}\)
\(A=7+2^3.7+2^6.7+2^9.7+....+2^{2016}.7+2^{2019}+2^{2020}\)
\(\text{Ta có:}2^{2019}+2^{2020}=8^{673}+8^{673}.2\equiv1+1.2\left(\text{mod 7}\right)\equiv3\left(\text{mod 7}\right)\Rightarrow A\text{ chia 7 dư 3}\)
CC
11 tháng 8 2019
1)Đặt A = \(\frac{1}{7.8}+\frac{1}{14.10}+\frac{1}{20.13}+...+\frac{1}{38.22}\)
\(\frac{1}{2}A=\frac{1}{8.14}+\frac{1}{14.20}+...+\frac{1}{38.44}\)
\(\frac{1}{2}A=6\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+...+\frac{1}{38}-\frac{1}{44}\right)\)
\(\frac{1}{2}A=6\left(\frac{1}{8}-\frac{1}{44}\right)\)
\(\frac{1}{2}A=6.\frac{9}{88}\)
\(\frac{1}{2}A=\frac{27}{44}\)
\(A=\frac{1}{44}:\frac{1}{2}=\frac{1}{22}\)
CC
11 tháng 8 2019
2) a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{3}-\frac{1}{9}-\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{2}{9}-\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{x\left(x+2\right)}=\frac{2}{9}-\frac{2018}{2020}\)
Hình như đề sai . Hoặc là mình sai >:
TM
7
9 tháng 1 2018
a ) 4 . ( x2 + 1 ) = 0
x2 + 1 = 0 : 4
x2 + 1 = 0
x2 = 0 - 1
x2 = - 1
x2 = - 12 => x = - 1
Vậy x = - 1
R
24 tháng 8 2019
Hello bạn, mk cx tên Mai nek.
\(\frac{2}{5}.\left(x-1\right)+1=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}\left(x+1\right)=\frac{3}{5}-1\)
\(\Rightarrow\frac{2}{5}\left(x+1\right)=-\frac{2}{5}\)
\(\Rightarrow x+1=-\frac{2}{5}:\frac{2}{5}\)
\(\Rightarrow x+1=-1\)
\(\Rightarrow x=-1-1\)
\(\Rightarrow x=-2\)
24 tháng 8 2019
\(\left(\frac{2}{7}\times x+1\right)\times\left(3-\frac{1}{2}\times x\right)=0\)
\(TH1:\frac{2}{7}\times x+1=0\)
\(\frac{2}{7}\times x=-1\)
\(x=-\frac{2}{7}\)
\(TH2:3-\frac{1}{2}\times x=0\)
\(\frac{1}{2}\times x=3\)
\(x=\frac{3}{2}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{7}\right\}\)