`4(x-5)-2^3=2^{4}.3`

`=>4(x-5)-8=16.3=48`

`=>4(x-5)=48+8=56`

`=>x-5=56:4=14`

`=>x=19`

Vậy `x=19`

4.(x-5)-2³=2⁴.3

4.(x-5)-8=16.3

4.(x-5)-8=48

4.(x-5)=48+8

4.(x-5)=56

x-5=56:4

x-5=14

x=14+5

x=19

     4 (x - 5) - 2³ = 2⁴*3

=> 4 (x - 5) - 2³ = 16*3 = 48

=> 4 (x - 5) = 48 + 2³ = 48 + 8 = 56

=> x - 5 = 56/4 = 14

=> x = 14 + 5 = 19

 Vậy x = 19

5(x+7)-10=2³.5

5(x+7)-10=8.5

5(x+7)-10=45 

5(x+7)=55  (45+10)

x+7=

Ta có: \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Leftrightarrow5\left(x+7\right)=50\)

\(\Leftrightarrow x+7=10\)

hay x=3

5(x+7)-10=2³.5

5(x+7)-10=8.5

5(x+7)-10=45

5(x+7)=55

x+7=\(\dfrac{55}{5}\)=11  =>x=11-7=3

Ta có: \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Leftrightarrow5\left(x+7\right)=50\)

\(\Leftrightarrow x+7=10\)

hay x=3

a) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)-10=40\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow x+7=\dfrac{50}{5}\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

b) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x-18=81\)

\(\Rightarrow9x=81+18\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

c) \(5^{25}\cdot5^{x-1}=5^{25}\)

\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)

\(\Rightarrow5^{x-1}=1\)

\(\Rightarrow5^{x-1}=5^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

a) 5(�+7)−10=23⋅55(x+7)−10=23⋅5

$\Rightarrow 5\left(x+7\right)-10=40$

$\Rightarrow 5\left(x+7\right)=40+10$

⇒�+7=505⇒x+7=550​

$\Rightarrow x+7=10$

$\Rightarrow x=10-7$

$\Rightarrow x=3$

b) 9�−2⋅32=349x−2⋅32=34

$\Rightarrow 9x-18=81$

$\Rightarrow 9x=81+18$

$\Rightarrow 9x=99$

⇒�=999⇒x=999​

$\Rightarrow x=11$

c) 525⋅5�−1=525525⋅5x−1=525

⇒5�−1=525:525⇒5x−1=525:525

⇒5�−1=1⇒5x−1=1

⇒5�−1=50⇒5x−1=50

$\Rightarrow x-1=0$

$\Rightarrow x=1$

a: x chia hết cho 4;5;10

nên \(x\in BC\left(4;5;10\right)\)

mà 10<=x<50

nên x=40

b: x=33

`@` `\text {Ans}`

`\downarrow`

`2^x * 4 = 128`

`=> 2^x = 128 * 4`

`=> 2^x = 512`

`=> 2^x = 2^9`

`=> x = 9`

Vậy, `x = 9`

`x^15 = x`

`=> x^15 - x = 0`

`=> x(x^14 - 1) = 0`

`=>` TH1: `x = 0`

`TH2: x^14 - 1 = 0`

`=> x^14 = 1`

`=> x = 1`

Vậy, `x \in {0; 1}`

`(2x+1)^3 = 125`

`=> (2x+1)^3 = 5^3`

`=> 2x + 1 = 5`

`=> 2x = 5 - 1`

`=> 2x =4`

`=> x = 4 \div 2`

`=> x = 2`

Vậy,` x = 2.`

`(x - 5)^4 = (x-5)^6`

`=> (x-5)^4 - (x-5)^6 = 0`

`=> (x-5)^4 * [ 1 - (x-5)^2] = 0`

`=> - (x-6)(x-5)^4(x-4) = 0`

`TH1: (x - 5)^4 = 0`

`=> x - 5 = 0`

`=> x = 0 +5`

`=> x = 5`

`TH2: x - 6=0`

`=> x=6`

`TH3: x-4=0`

`=> x = 4`

Vậy, `x \in {4; 5; 6}`

a: =>2^x=32

=>x=5

b: =>x^15-x=0

=>x(x^14-1)=0

=>x=0; x=1;x=-1

c: =>2x+1=5

=>2x=4

=>x=2

d: =>(x-5)^4[(x-5)^2-1]=0

=>(x-5)(x-4)(x-6)=0

=>x=5;x=4;x=6

123 – 5(x + 4) = 38

5(x + 4) = 123 – 38

5(x + 4) = 85

x + 4 = 85 : 5

x + 4 = 17

x = 17 – 4

x = 13

Vậy x = 13

x<7/12:4/5

=>x<35/16

Ta có: 35/16>35/35=1;35/16>32/16=2

=> x=1;x=2