a) \(1=\left(2x+0,5\right)^{600}\)

\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)

\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)

b) \(\left(x-0,125\right)^2=0,25\)

\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)

\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)

c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)

\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)

\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`1 = (2x + 0,5)^600`

`=> (2x+0,5)^600 = (+-1)^600`

`=> \text {TH1: } 2x + 0,5 = 1`

`=> 2x = 1 - 0,5`

`=> 2x = 0,5`

`=> x = 0,5 \div 2`

`=> x = 0,25`

`\text {TH2: } 2x + 0,5 = -1`

`=> 2x = -1 - 0,5`

`=> 2x = -1,5`

`=> x = -1,5 \div 2`

`=> x = -0,75`

Vậy, `x \in {-0,75; 0,25}.`

`b)`

`(x - 0,125)^2 = 0,25`

`=> (x - 0,125)^2 = (+-0,5)^2`

`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)

Vậy, `x \in {-0,375; 0,625}.`

`c)`

`(x - 3)^11 = (x - 3)^41`

`=> (x - 3)^11 - (x - 3)^41 = 0`

`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`

`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy, `x \in {3; 4}.`

a, 1=(2x+0,5)⁶⁰⁰

=>1⁶⁰⁰=(2x+0,5)⁶⁰⁰

=>1=2x+0,5

=>2x+0,5=1

=> 2x= 1+0,5

=>2x= 1,5

=:>x= 1,5:2

=>x=0,75

b, (x-0,125)²=0,25

=>(x-0,125)²=(0,5)²

=>x-0,125=0,5

=>x=0,5 +0,125

=>x= 0,625

c, (x-3)¹¹=(x-3)⁴¹

(x-3)¹¹ -(x-3)⁴¹=0

(x-3)¹¹-(x-3)¹¹ .(x-3)³⁰=0

(x-3)¹¹.[1-(x-3)^{30 ]}=0

(x-3)¹¹.(4-x)³⁰=0

(x-3)¹¹=0 hoặc (4-x)³⁰=0

(x-3)¹¹=0          ( 4-x)³⁰=0

x-3=0                4-x=0

x=0+3                  x=4-0

x=3                     x=4

vậy xϵ{3,4}

a: x=2/3-4/5=10/15-12/15=-2/15

b: 1/2-x=7/12

=>x=1/2-7/12=-1/12

c: =>7/2:x=-7/2

=>x=-1

d: =>1/6x=3/8-5/2=3/8-20/8=-17/8

=>x=-17/8*6=-102/8=-51/4

e: =>1,5x=-1,5

=>x=-1

Bài 2: 

a: 7/13:x=14/39

nên \(x=\dfrac{7}{13}:\dfrac{14}{39}=\dfrac{7}{13}\cdot\dfrac{39}{14}=\dfrac{3}{2}\)

b: x+3/7=2/5+3/10=4/10+3/10=7/10

=>x=7/10-3/7=49/70-30/70=19/70

1)\(x+0,5+x+1,5+x+2,5=33\)

\(\Leftrightarrow3x=33-0,5-1,5-2,5=28,5\)

\(\Leftrightarrow x=9,5\)

2)\(\left(x+0,9\right)\left(1-0,4\right)=2412\)

\(\Leftrightarrow\left(x+0,9\right)\cdot0,6=2412\)

\(\Leftrightarrow x+0,9=4020\)

\(\Leftrightarrow x=1019,1\)

fffffff

4028  x  0,5 + 4028 + 2014 : \(\dfrac{1}{2}\) x 1,5 + 4028 : 0,5

= 4028 x 0,5 + 4028 x 1  + 2014 x 2 x 1,5 + 4028 x 2

= 4028 x 0,5 + 4028 x 1 + 4028 x 1,5 + 4028 x 2

= 4028 x ( 0,5 + 1 + 1,5 + 2)

= 4028 x 5

= 20140

\(\text{a, (x - 1)^2 = 1}\\ x-1=1\\ x=2\)

\(x-1=1\Rightarrow x=2\)

\(x+\dfrac{2}{-15}=\dfrac{5}{3}\)

\(x=\dfrac{9}{5}\)

\(x+\dfrac{2}{-15}=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}-\dfrac{-2}{15}\)

\(x=\dfrac{-23}{15}\)

Lời giải:

$x+\frac{2}{-15}=\frac{-5}{3}$

$x=\frac{-5}{3}-\frac{2}{-15}=\frac{-5}{3}+\frac{2}{15}$

$x=\frac{-23}{15}$

\(x=\dfrac{1}{8}\)