a: \(2^x=2^3\)

nên x=3

c: \(11^x=1331\)

nên x=3

d: \(2^x+4=12\)

nên \(2^x=8\)

hay x=3

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

\(a,\Rightarrow x^2+4x+25-x^2=3\\ \Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\\ b,\Rightarrow\left(2x-3-4x-3\right)\left(2x-3+4x+3\right)=0\\ \Rightarrow6x\left(-2x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)

a) x + \(\dfrac{3}{4}\) = \(\dfrac{5}{3}\)

    x = \(\dfrac{5}{3}\) - \(\dfrac{3}{4}\)

    x = \(\dfrac{20}{12}\) - \(\dfrac{9}{12}\)  

    x =  \(\dfrac{11}{12}\) 

b) x - \(\dfrac{2}{3}\) = \(\dfrac{7}{2}\)

    x = \(\dfrac{7}{2}\) + \(\dfrac{2}{3}\) 

    x = \(\dfrac{21}{6}\) + \(\dfrac{4}{6}\) 

    x = \(\dfrac{25}{6}\)

nhanh lên nhé 

sos

chi chi mà cả nhả

b: Ta có: 7x-8=713

nên 7x=721

hay x=103

c: Ta có: x-36:18=12

nên x-2=12

hay x=14

d: Ta có: (x-36):18=12

nên x-36=216

hay x=252

a.      124 + (118 – x) = 217
                     118 - x  = 93
                            - x = - 25
                              x  = 25

b.      7x – 8 = 713
          7x      =721
            x      =103

c.       x- 36:18 = 12
          x- 2        =12
          x             =14

d.      (x- 36):18 = 12
           x-36        = 216
           x             = 252
          

e.      (x-47) -115 = 0
          x-47          = 115
          x               = 162

f.        315 + (146 – x) = 401
                      146-x     =86
                            -x     = -60
                             x     = 60

g.      (6x – 39 ) : 3 = 201
           6x - 39        = 603
           6x               = 642
             x               = 107

h.      23 + 3x = 5⁶ : 5³
         23 +3x   = 125
                3x    = 102
                  x=34

\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)

\(a,x.4,25=7\\ \Rightarrow x=7:4,25\\ \Rightarrow x=\dfrac{28}{17}\\ b,116:x=14,5\\ \Rightarrow x=116:14,5\\ \Rightarrow x=8\\ c,15.x=8\\ \Rightarrow x=\dfrac{8}{15}\\ c,729:x=36\\ \Rightarrow x=729:36\\ \Rightarrow x=20,25\)

Tìm x:

a) x . 4,25 = 17 

x = 17 : 4,25

x = 4

b) 116 : x = 14,5

x = 116 : 14,5

x = 8

c) 15 . x = 8 

x = 8 : 15

x = 0.53333333333

d) 729 : x = 36

x = 729 : 36

x = 20.25

a: =>x-5=9

=>x=14

b: căn x-10=-2

=>\(x\in\varnothing\)

c: căn 2x-1=căn 5

=>2x-1=5

=>2x=6

=>x=3

d: căn 4-5x=12

=>4-5x=144

=>5x=-140

=>x=-28

e: =>7|x-1|=35

=>|x-1|=5

=>x-1=5 hoặc x-1=-5

=>x=6 hoặc x=-4

f: =>\(\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)

=>x+3=0 hoặc x-3=25

=>x=28 hoặc x=-3

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}