\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)

a) \(8x+56:14=60\)

\(\Rightarrow8x+4=60\)

\(\Rightarrow8x=56\)

\(\Rightarrow x=\dfrac{56}{8}\)

\(\Rightarrow x=7\)

b) Mình làm rồi nhé !

c) \(41-2^{x+1}=9\)

\(\Rightarrow2^{x+1}=41-9\)

\(\Rightarrow2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

d) \(3^{2x-4}-x^0=8\)

\(\Rightarrow3^{2x-4}-1=8\)

\(\Rightarrow3^{2x-4}=9\)

\(\Rightarrow3^{2x-4}=3^2\)

\(\Rightarrow2x-4=2\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

g) \(65-4^{x+2}=2014^0\)

\(\Rightarrow65-4^{x+2}=1\)

\(\Rightarrow4^{x+2}=64\)

\(\Rightarrow4^{x+2}=4^3\)

\(\Rightarrow x+2=3\)

\(\Rightarrow x=1\)

i) \(120+2\left(4x-17\right)=214\)

\(\Rightarrow2\left(4x-17\right)=214-120\)

\(\Rightarrow2\left(4x-17\right)=94\)

\(\Rightarrow4x-17=47\)

\(\Rightarrow4x=47+17\)

\(\Rightarrow4x=64\)

\(\Rightarrow x=16\)

a: \(8x+56:14=60\)

=>8x+4=60

=>8x=60-4=56

=>x=56/8=7

b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)

=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)

=>2x-3=3

=>2x=6

=>x=3

c: \(41-2^{x+1}=9\)

=>\(2^{x+1}=41-9=32\)

=>x+1=5

=>x=4

d: \(3^{2x-4}-x^0=8\)

=>\(3^{2x-4}-1=8\)

=>\(3^{2x-4}=8+1=9\)

=>2x-4=2

=>2x=6

=>x=3

g: \(65-4^{x+2}=2014^0\)

=>\(65-4^{x+2}=1\)

=>\(4^{x+2}=65-1=64\)

=>x+2=3

=>x=1

i: 120+2(4x-17)=214

=>2(4x-17)=214-120=94

=>4x-17=94/2=47

=>4x=64

=>\(x=\dfrac{64}{4}=16\)

b) 9x^2-(x-4)^2+0

Ta có: \(100< 5^{2x-1}\le5^6\)

\(\Leftrightarrow5^2< 5^{2x-1}\le5^6\)

\(\Leftrightarrow2x-1\in\left\{3;5\right\}\)

\(\Leftrightarrow2x\in\left\{4;6\right\}\)

hay \(x\in\left\{2;3\right\}\)

a: 7x+58=100

nên 7x=42

hay x=6

c: x-56:x=16

nên x-14=16

hay x=30

c)x - 56 : 4  = 16

x - 56       = 16 : 4 

x- 56        = 4

x               =4 + 56

x               = 60 

d)101 + (36 - 4x) = 105 

          (36- 4x ) = 105 - 101 

          36 - 4x = 4

                4x = 36 - 4 

                4x = 32

                  x = 32:4

                  x = 8

a) 58 + 7x = 100

            7x = 100 - 58

            7x = 42

              x = 42 : 7

              x = 6

b) (x - 12) : 12 = 12

              x - 12 = 12 . 12

             x - 12 = 144

                    x = 144 + 12

                    x = 156

Ta có: 100 < 52x – 1 < 56

=> 52 < 100 < 52x-1 < 56

=> 2 < 2x – 1 < 6

=> 2 + 1 < 2x < 6 + 1

=> 3 < 2x < 7

Vì x ∈ N nên suy ra: x ∈ {2; 3} là thỏa mãn.

Ta có 100=5².4

\(\Rightarrow5^3\le5^{2x-1}< 5^6\)

\(\Rightarrow3\le2x-1< 6\)

\(\Rightarrow4\le2x< 7\)

\(\Rightarrow2\le x< 3,5\)

Mà \(x\) là số tự nhiên

\(\Rightarrow x=2\) hoặc \(x=3\)

x ϵ {2;3}

x15+32x+x1=780

xxx+(15+320+1)=780

xxx+336    =780

xxx           = 780-336  

xxx             = 444

<=>x.(15+32+1)=780

<=>x.48           =780

<=>x                =780:48

<=>x                =16,25

Vậy x=15,25

cá này vốn đã ko phải toán lớp 3 maK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!