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=12

b: x=-19

c)x=-25

d)x=30

\(a,\sqrt{x}=3\Leftrightarrow x=9\\ b,\sqrt{x}=6\Leftrightarrow x=36\\ c,\sqrt{x}=8\Leftrightarrow x=64\\ d,\sqrt{x}=12\Leftrightarrow x=144\\ e,2\sqrt{x}=10\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\\ f,3\sqrt{x}=21\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\\ g,\sqrt{x}=8\Leftrightarrow x=64\\ h,2+\sqrt{x}=11\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

a. \(\sqrt{x}=3\)

<=> \(\left(\sqrt{\sqrt{x}}\right)^2-\left(\sqrt{3}\right)^2=0\)

<=> \(\left(\sqrt{\sqrt{x}}-\sqrt{3}\right)\left(\sqrt{\sqrt{x}}+\sqrt{3}\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt{\sqrt{x}}-\sqrt{3}=0\\\sqrt{\sqrt{x}}+\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\left(Vnghiêm\right)\end{matrix}\right.\)

Vậy nghiệm của PT là S = \(\left\{9\right\}\)

a) \(\sqrt{x}=3\Leftrightarrow x=9\)

b) \(\sqrt{x}=6\Leftrightarrow x=36\)

c) \(\sqrt{x}=8\Leftrightarrow x=64\)

d) \(\sqrt{x}=12\Leftrightarrow x=144\)

e) \(2\sqrt{x}=10\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\)

f) \(3\sqrt{x}=21\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)

g) \(\sqrt{x}=8\Leftrightarrow x=64\)

h) \(2+\sqrt{x}=11\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

\(a,x-\dfrac{1}{4}=\dfrac{7}{2}.\dfrac{-3}{5}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{-21}{10}\\ \Rightarrow x=\dfrac{-21}{10}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-37}{20}\\ b,\dfrac{x}{134}=\dfrac{9}{7}.\dfrac{5}{-11}\\ \Rightarrow\dfrac{x}{134}=\dfrac{-45}{77}\\ \Rightarrow x=\dfrac{-45}{77}.134\\ \Rightarrow x=\dfrac{-6030}{77}\)

\(x-\dfrac{1}{4}=\dfrac{7}{2}\cdot\dfrac{-3}{5}\)

\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)

\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)

\(x=\dfrac{-37}{20}\)

b ) \(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{5}{-11}\)

      \(\dfrac{x}{134}=\dfrac{9}{7}\cdot\dfrac{-5}{11}\)

       \(\dfrac{x}{134}=\dfrac{-45}{77}\)

       \(x=\dfrac{-45}{77}\cdot134\)

       \(x=-\dfrac{6034}{77}\)

a.\(\dfrac{1}{3}\) + x  = \(\dfrac{5}{6}\)

       x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)

      x = \(\dfrac{1}{2}\)

b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\) 

   | x-1|        = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)

  |x-1|        = \(\dfrac{3}{2}\)

\(\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{3}{2}+1\\x=-\dfrac{3}{2}+1\end{matrix}\right.\)

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

c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1

            \(\dfrac{2}{3}\) (\(\dfrac{x}{2}\) + 3) = 1 - \(\dfrac{1}{3}\)

             \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)

                   \(\dfrac{x}{2}\) + 3 = 1

                   \(\dfrac{x}{2}\)       = 1 - 3

                    \(\dfrac{x}{2}\)    = -2

                     \(x\) = -4

d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)

(x+2)² = 27.3

(x+2) =9²

\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)

\(\left[2040:\left(47-3x\right)+33\right].11=1683\)

\(\Leftrightarrow2040:\left(47-3x\right)+33=153\)

\(\Leftrightarrow2040\left(47-3x\right)=120\)

\(\Leftrightarrow47-3x=17\)

\(\Leftrightarrow3x=30\Leftrightarrow x=10\)

a: Ta có: \(71-\left(x+33\right)=26\)

\(\Leftrightarrow x+33=45\)

hay x=12

b: Ta có: \(\left(x+73\right)-26=76\)

\(\Leftrightarrow x+73=102\)

hay x=29

c: Ta có: \(45-\left(x+9\right)=6\)

\(\Leftrightarrow x+9=39\)

hay x=30

a) 71 - ( 33 + x ) = 26

(33 + x ) = 71 -26

33 + x = 45

x        = 45 - 33 = 12

b) ( x + 73 ) - 26 = 76

    ( x + 73 )        = 102

    x                     = 102 - 73 = 29

c) 45 - ( x + 9 ) = 6

 ( x + 9 )           = 45 - 6 = 39

 x                      = 39 - 9 = 30

d) 89 - ( 73 - x ) = 20 

 73 - x  = 89 - 20

73 - x = 69

x          = 73 - 69 = 4

e) 4(x+41) = 400

   x + 41 = 400 : 4 = 100

   x = 100 - 41 = 59

f) 11(x-9 ) = 77

     x - 9 = 77 : 11 = 7

    x = 7 + 9 = 16

g) x + 7 = 25 + 13 = 38

    x = 38 - 7 = 31

h) x + 4 = 198 - 120 = 78

   x = 78 - 4 = 74

i) x - 9 = 350 : 5 = 70

   x = 70 + 9 = 79

j) 2x - 49 = 5 . 9 = 45

   2x = 45 + 49 = 94

   x = 94 : 2 = 47

k) 25 + 3( x - 8 ) = 106

    3(x-8 ) = 106 - 25 = 81

   x - 8 = 81 : 3 = 27 

x = 27 +8= 35

l) 9( x + 4 ) - 25 = 20

   9( x + 4 ) = 20 + 25 = 45

     x + 4 =45 : 9 = 5

   x = 5 - 4 = 1

m) 200 - ( 2x + 6 ) = 64

     2x + 6 = 200 - 64 = 136

   2x = 136 - 6 = 130

x = 130 : 2 = 65

n) 2(x-51) = 16 + 20 = 36

   x - 51 = 36 : 2 = 13

  x = 13 + 51 = 64

o) 450 : ( x - 19 ) = 50

  x - 19 = 450 : 50 = 9

x = 19 + 9 = 28

p)4(x-3) = 49 - 1 = 48

x - 3 = 48 : 4 = 12

x + 15

q) 156 - ( x+61 ) = 82

x + 61 = 156 - 82 = 74

x = 74 - 61 = 13

r) ( x - 35 ) - 120 = 0

x - 35 = 120

x = 120 + 35 = 155

1.a) 4.(12+88) = 4.100=400

b)7²- 6 .[2.8:8 ]=7²-6.[16:8]=7²-6 .2= 49-12=37  ( tui ko chắc lắm )

2.a) x=26-11=15 

b) (4x-15)³=5³ 

4x-15=5 

4x= 5+15

4x=20

vậy x = 20

Bài 2: 

a: Ta có: x+11=26

nên x=15

b: Ta có: \(\left(4x-15\right)^3=125\)

\(\Leftrightarrow4x-15=5\)

hay x=5