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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
a)5x+1=125
=>5x+1=53
=>x+1=3
=>x=2
vậy x=2
b)42x+1=64
=>42x+1=43
=>2x+1=3
=>x=1
vậy x =1
e)=>43x+2017=42020-3
=>3x+2017=2017
=>x=0
vậy x=0
f)=>2x+2x x 23=144
=>2x x (1+23)=144
=>2x x 9=144
=>2x=16
=>2x=24
=>x=4
vậy x=4
a)(4-x)2+15=40
(4-x)2=40-15
(4-x)2=25
(4-x)2=52
=>4-x=5
x=5-4
x=1
a) ( 4 - x )2 + 15 = 40
( 4 - x )2 = 40 - 15
( 4 - x )2 = 25
( 4 - x )2 = 52
4 - x = 5
x = 4 - 5
x = ( - 1 )
b, ( 3x - 2 )10 = ( 3x - 2 )4
=> x \(\in\left\{0;1\right\}\)
c, X = 1 + 4 + 42 + ......... + 42017
4X = 4 + 41 + ......... + 42018
4X - X = ( 4 + 41 + ........ + 42018 ) - ( 1 + 4 + 42 + ......... + 42017 )
4X - X = 4 + 41 + ......... + 42018 - 1 - 4 - 42 - ......... - 42017
=> 3X = 42018 - 1
=> X = \(\frac{4^{2018}-1}{3}\)
+) \(A=3\left(x-4\right)^4-4\ge-4\)
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 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\2x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\2x=y\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
1) x - 36 + 12 = - x+ 10
=> x + x = 10 + 24
=> 2x = 34
=> x = 34/2 = 17
2) (x + 15) - (11 - x) = (-2)2
=> x + 15 - 11 + x = 4
=> 2x = 4 - 4
=> 2x = 0
=> x = 0
3) 40 - 4x2 = (-6)2
=> 40 - 4x2 = 36
=> 4x2 = 40 - 36
=> 4x2 = 4
=> x2 = 1
=> x = \(\pm\)1
4) (-50) + 10x2 = (-25) x |-2|
=> -50 + 10x2 = -50
=> 10x2 = -50 + 50
=> 10x2 = 0
=> x2 = 0
=> x = 0
5) |x + 1| = 2020
=> \(\orbr{\begin{cases}x+1=2020\\x+1=-2020\end{cases}}\)
=> \(\orbr{\begin{cases}x=2019\\x=-2021\end{cases}}\)
6) (x + 1)5 + 8 = 0 (xem lại đề)
7) (-20) + x3 : 16 = -24
=> x3 : 16 = -24 + 20
=> x3 : 16 = -4
=> x3 = -4 . 16
=> x3 = -64 = (-4)3
=> x = -4
9) x14 = x17
=> x14 - x17 = 0
=> x14(1 - x3) = 0
=> \(\orbr{\begin{cases}x^{14}=0\\1-x^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
10) (-36) + (1 - x)2 = 0
=> (1 - x)2 = 36
=> (1 - x)2 = 62
=> \(\orbr{\begin{cases}1-x=6\\1-x=-6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
a) \(20\cdot2^x+1=10\cdot4^2+1\)
\(\Leftrightarrow2\cdot10\cdot2^x=10\cdot4^2\)
\(\Leftrightarrow10\cdot2^{x+1}=10\cdot2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b) \(\left(4-\frac{x}{2}\right)^3-1=2\cdot\left(2^3-\frac{5}{2^0}\right)+1\)
\(\Leftrightarrow\left(4-\frac{x}{2}\right)^3=2\cdot3+1+1\)
\(\Leftrightarrow\left(4-\frac{x}{2}\right)^3=8=2^3\)
\(\Rightarrow4-\frac{x}{2}=2\)
\(\Leftrightarrow\frac{x}{2}=2\)
\(\Rightarrow x=4\)
\(2^5.3^2+2^5.11-2^6.5-5\)
\(=2^5.9+2^5.11-2^5.2^1.5-5\)
\(=2^5.\left(9+11-2.5\right)-5\)
\(=32.\left(9+11-10\right)-5\)
\(=32.10-5\)
\(=320-5\)
\(=315\)
\(-2017-\left[\left(15-2017\right)+\left(-115\right)\right]\)
\(=-2017-\left[\left(-2002\right)+\left(-115\right)\right]\)
\(=-2017-\left(-2117\right)\)
\(=-2017+2117\)
\(=100\)
Vì 24 chia hết cho x, 120 chia hết cho x và 10<x<20 nên x ƯC(24,120)
Ta có : 24 =12.2 ; 120= 10.12
ƯCLN(24,120) = 12
Mà Ư(12) = { 1,2,3,4,6,12}
=>ƯC(24,120) = { 1,2,3,4,6,12}
Vì 10<x<20
=> x = 12
Vậy x = 12
3.|x-1| = 28:23 + 20170
3.|x-1| = 25+ 1
3.|x-1| = 32 + 1
3.|x-1| = 33
|x-1| = 33 : 3
|x-1| =11
=> x -1 =11
x = 11+1
x = 22