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a) (x - 2)^2 = 16
(x - 2)^2 = 4^2 = (-4)^2
=> x - 2 = 4 hoặc x - 2 = -4
- Nếu x - 2 = 4
=> x = 4 + 2 = 6
- Nếu x - 2 = -4
=> x = (-4) + 2 = -2
Vậy x = 6 hoặc x = -2
b) (2x - 1)^3 = 8
(2x - 1)^3 = 2^3
=>2x - 1 = 2
2x = 2 + 1 = 3
\(x=\frac{3}{2}\)
c) (2x - 1)^4 = 81
(2x - 1)^4 = 3^4=(-3)^4
=> 2x - 1 = 3 hoặc 2x - 1 = -3
- Nếu 2x - 1 = 3
=>2x = 3 + 1 = 4
=>x = 4 : 2 = 2
- Nếu 2x - 1 = -3
2x = (-3) + 1 = -2
x = -1
Vậy x = 2 hoặc x = -1
1a) x - 7 = -18
=> x = -18 + 7
=> x = -11
b) -2x + 8 = -24
=> -2x = -24 - 8
=> -2x = -32
=> x = -32 : (-2)
=> x = 16
c) 4 - 2x = 59
=> 2x = 4 - 59
=> 2x = -55
=> x = -55 : 2
=> x = -27,5
d) -36 - x = 59
=> x = -36 - 59
=> x = -95
k) -81 : x = (-3)
=> x = -81 : (-3)
=> x = 27
g) -16 : x = 2
=> x = -16 : 2
=> x = -8
m) (6 - 27) - (3 - x) = (-7) . (-8)
=> -21 - (3 - x) = 56
=> 3 - x = -21 - 56
=> 3 - x = -77
=> x = 3 + 77
=> x = 80
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
a. (2x-1)4=81
=>(2x-1)4=34
=>2x-1=3
=>2x=3+1
=>2x=4
=>x=4:2
=>x=2
b.(x-1)5=-32
=>(x-1)5=(-2)5
=>x-1=-2
=>x=-2+1
=>x=-1
c.(2x-1)6=(2x-1)8
mà chỉ có: (-1)6=(-1)8; 06=08; 16=18
=> để (2x-1) \(\in\){-1;0;1} thì x \(\in\){0; 1/2; 1}
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
c: =>2x+1=4 hoặc 2x+1=-4
=>x=3/2 hoặc x=-5/2
h: =>x-3=4
=>x=7
g: =>2x-1=3
=>2x=4
=>x=2
f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)
=>x*-5/6=5/6
=>x=-1
d: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>x=-1 hoặc x=2