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Ta có: \(x^2\ge0;\left|x+y\right|\ge0;\forall x,y\)
=> \(M=2015+3\left(x^2+1\right)^{2016}+\left|x+y\right|^{2017}\)
\(\ge2015+3\left(0+1\right)^{2016}+0^{2017}=2018\)
Dấu "=" xảy ra khi và chỉ khi: \(\hept{\begin{cases}x^2=0\\\left|x+y\right|=0\end{cases}\Leftrightarrow x=y=0}\)
Vậy gtnn của M = 2018 đạt tại x = y = 0.
a) \(3^{x+1}=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
b) \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\left(\frac{1}{2}\right)^6\)
\(\Leftrightarrow x+1=6\Leftrightarrow x=5\)
c) \(\frac{81}{3x}=9\)
\(\Leftrightarrow3x=9\Leftrightarrow x=3\)
d) \(2^{x+1}+2^{x+2}=192\)
\(\Leftrightarrow2^x.2+2^x.4=192\)
\(\Leftrightarrow2^x.6=192\Leftrightarrow2^x=32\Leftrightarrow x=5\)
e) Ta có : \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}\Rightarrow\left(x-1\right)^{2020}+\left(y+2\right)^{2020}\ge0}\)
Mà \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Bài giải
a, \(3^{x+1}=243\)
\(3^{x+1}=3^5\)
\(\Rightarrow\text{ }x+1=5\)
\(\Rightarrow\text{ }x=4\)
b, \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\frac{1}{2^{x+1}}=\frac{1}{2^6}\)
\(2^{x+1}=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
c, \(\frac{81}{3x}=9\)
\(27x=81\)
\(x=3\)
d, \(2^{x+1}+2^{x+2}=192\)
\(2^{x+1}\left(1+2\right)=192\)
\(2^{x+1}\cdot3=192\)
\(2^{x+1}=64=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
e, \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
Mà \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}}\) với mọi x,y nên \(\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
\(\Rightarrow\text{ }x=1\text{ ; }y=-2\)
C với D mình làm sau vì nó phức tạp hơn ... E với F trước nhé
E = | 3x + 1 | + 2| x - y | + 1
\(\hept{\begin{cases}\left|3x+1\right|\ge0\\2\left|x-y\right|\ge0\end{cases}\forall}x,y\Rightarrow\left|3x+1\right|+2\left|x-y\right|+1\ge1\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x+1=0\\x-y=0\end{cases}}\Leftrightarrow x=y=-\frac{1}{3}\)
=> MinE = 1 <=> x = y = -1/3
F = 5| x - 1 | + 1/2| 2x + y | + 2020
\(\hept{\begin{cases}5\left|x-1\right|\ge0\\\frac{1}{2}\left|2x+y\right|\ge0\end{cases}\forall}x,y\Rightarrow5\left|x-1\right|+\frac{1}{2}\left|2x+y\right|+2020\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\2x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
=> MinF = 2020 <=> x = 1 ; y = -2
C = 2| x - 1 | + | 2x + 3 | - 2020
= | 2x - 2 | + | 2x + 3 | - 2020
= | 2x - 2 | + | -( 2x + 3 ) | - 2020
= | 2x - 2 | + | -2x - 3 | - 2020
Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :
C = | 2x - 2 | + | -2x - 3 | - 2020 ≥ | 2x - 2 - 2x - 3 | - 2020 = | -5 | - 2020 = 5 - 2020 = -2015
Dấu "=" xảy ra khi ab ≥ 0
=> ( 2x - 2 )( -2x - 3 ) ≥ 0
Xét hai trường hợp :
1. \(\hept{\begin{cases}2x-2\ge0\\-2x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge2\\-2x\ge3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le-\frac{3}{2}\end{cases}}\)( loại )
2. \(\hept{\begin{cases}2x-2\le0\\-2x-3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le2\\-2x\le3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-\frac{3}{2}\end{cases}}\Leftrightarrow-\frac{3}{2}\le x\le1\)
=> MinC = -2015 <=> \(-\frac{3}{2}\le x\le1\)
D = | 3 - 2x | + 2| 1 - x | + 1/2
= | 3 - 2x | + | 2 - 2x | + 1/2
= | -( 3 - 2x ) | + | 2 - 2x | + 1/2
= | 2x - 3 | + | 2 - 2x | + 1/2
Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :
D = | 2x - 3 | + | 2 - 2x | + 1/2 ≥ | 2x - 3 + 2 - 2x | + 1/2 = | -1 | + 1/2 = 1 + 1/2 = 3/2
Dấu "=" xảy ra khi ab ≥ 0
=> ( 2x - 3 )( 2 - 2x ) ≥ 0
Xét hai trường hợp :
1. \(\hept{\begin{cases}2x-3\ge0\\2-2x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge3\\-2x\ge-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( loại )
2. \(\hept{\begin{cases}2x-3\le0\\2-2x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le3\\-2x\le-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)
=> MinD = 3/2 <=> \(1\le x\le\frac{3}{2}\)
\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)
= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)
= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)
= \(\dfrac{337.2023}{2}\)
= \(\dfrac{\text{681751}}{2}\)
|x-3|;|x+7| > 0
=>F > -111+0=-111
=>Fmin=-111
dấu "=" xảy ra<=>x=3;x=-7
cái thứ 2 có min đâu bạn ơi?,cả thứ 3 nữa
\(E=\left|x-1\right|+\left|x-9\right|\)
\(E=\left|x-1\right|+\left|9-x\right|\ge\left|x-1+9-x\right|=8\)
Min E = 8
\(\Leftrightarrow1\le x\le9\)