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\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
\(b,x^6=x^2\)
\(x^6-x^2=0\)
\(x^2\cdot\left(x^4-1\right)=0\)
\(\orbr{\begin{cases}x^2=0\\x^4-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(c\text{}\text{}\text{}\text{},\left(x-2\right)\cdot\left(x-5\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)
\(d,x^{10}-x^5=0\)
\(x^5\cdot\left(x^5-1\right)=0\)
\(\orbr{\begin{cases}x^5=0\\x^5=1\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
\(e,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\left(x-5\right)^4\cdot\left[1-\left(x-5\right)^2\right]=0\)
\(\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm1+5\end{cases}}}\)
\(\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
\(\left(2x+1\right)^3=125\Rightarrow\left(2x+1\right)^3==5^3\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1=4\Rightarrow x=4:2=2\)
\(x^6=x^2\Rightarrow x^2.x^4=x^2\)Vì vậy nên \(x=\pm1\)
\(\left(x-2\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\Rightarrow x=0+2=5\\x-5=0\Rightarrow X=0+5=5\end{cases}}\)

1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4

a, 311 - x + 82 = 46 + ( x -21 )
311 + 82 -x = 46 + x -21
393 - x = 25 + x
393 - 25 = x + x
368 = 2x
= > x = 184
b,−( x − 3 + 85 ) = ( x + 70 − 71 ) − 5
- x + 3 - 85 = x + 70 - 71 - 5
-x -82 = x - 6
-82 + 6 = x + x
-76 = 2x
x = -38
c,− 2 ( x + 6) + 6 ( x − 10 ) = 8
-2x - 12 + 6x - 60 = 8
-2x+ 6x -12 - 60 = 8
4x -72 = 8
4x = 80
x = 20
d,| x | = 41
=> x = 41 hoặc x = 41
e, | − 8 | . | x | = | −56 |
8 . | x | = 56
=> | x | = 7
=> x=7 hoặc x = -7
f , |x − 2 | + 2 x = 19 với x ≥ 2
Ta có : | x - 2 | + 2x = 19
| x -2 | = 19 - 2x
=> x - 2 = 19-2x hoặc x -2 = - ( 19-2x)
+) x -2 = 19-2x
=> x + 2x = 19 +2
=> x.(2+1 ) = 21
=> x . 3 = 21
=> x =7
+) x-2 = - ( 19 -2x )
=> x -2 = -19 +2x
=> -2 + 19 = 2x -x
=> 17 = x
Vậy x = 17 hoặc x = 7

1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna

a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).

\(a\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\\ =>\left(x-\frac{1}{2}\right)=\frac{1}{3}\\ =>x=\frac{1}{3}+\frac{1}{2}\\ =>x=\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\\ =>\left(x+\frac{1}{2}\right)=\frac{2}{5}\\ =>x=\frac{-1}{10}\)
d) (2x+3)2016=(2x+3)2018 khi 2x+3=0 hoặc 1
Nếu 2x+3=0
=2x=-3 ( loại )
Nếu 2x+3=1
=>2x=-2
=>x=-1 ( thỏa )

\(\left(2x-3\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=3\end{matrix}\right.\)
*\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{1}{2x-1}=-5-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{3\left(2x-1\right)}=\frac{-21}{4}\)
\(\Leftrightarrow-63\left(2x-1\right)=4\)
\(\Leftrightarrow2x-1=-\frac{4}{63}\)
\(\Leftrightarrow2x=\frac{59}{63}\)
\(x=\frac{59}{126}\)
\(\left[\left(2x-3\right).10\right]+6=66\)
\(\left(2x-3\right).10=66-6\)
\(\left(2x-3\right).10=60\Rightarrow2x-3=60:10\)
\(2x-3=6\Rightarrow2x=6-3\Rightarrow2x=3\)
\(x=3:2\Rightarrow x=\frac{3}{2}\)
[ ( 2x - 3 ) .10 ] + 6 = 66
( 2x - 3 ) . 10 = 66 - 6
( 2x - 3 ) . 10 = 60
2x - 3 = 60 : 10
2x - 3 = 6
2x = 3
x = 3 : 2
\(\Rightarrow x=\frac{3}{2}\)