a). 5^x = 125

=>5^x = 5³

=> x = 3

b) 3^2x = 81

=> 3^2x = 3⁴

=> 2x = 4

=> x = 2

c). 5^2x-3 – 2.5² = 5² .3

=>5^2x: 5³ = 5² .3 + 2.5²

=>5^2x: 5³ = 5² .5

=>
5^2x = 5² .5.5³

=>5^2x = 5⁶

=> 2x = 6

=> x=3

a) 5x = 125

x = 125 : 5

x = 25

b) 32x = 81

x = 81 : 32

x = 81/ 32

c) 52x - 3 - 2.52 = 52.3

52x - 3 - 104 = 156

x - 3 - 104 = 156 : 52

x - 3 - 104 = 3

x - 3 = 3 + 104

x - 3 = 107

x = 107 + 3

x = 110

Ko chắc lắm , sai thì Sorry nhé .

\(a,\\ 5x=125\\ < =>x=125:5\\ < =>x=25\)

\(b,\\ 32x=81\\ < =>x=81:32\\ < =>x=\frac{81}{32}\)

\(c,\\ 52x-3-2.52=52.3\\ < =>52x-3-104=156\\ < =>52x-107=156\\ < =>52x=156+107\\ < =>52x=263\\ < =>x=x=\frac{263}{52}\)

a: Ta có: \(2^{x-1}=32\)

\(\Leftrightarrow x-1=5\)

hay x=6

b: Ta có: \(3^{2x+1}=81\)

\(\Leftrightarrow2x+1=4\)

\(\Leftrightarrow2x=3\)

hay \(x=\dfrac{3}{2}\)

c: Ta có: \(2^x-26=6\)

\(\Leftrightarrow2^x=32\)

hay x=5

d: Ta có: \(27\cdot3^x=243\)

\(\Leftrightarrow3^x=9\)

hay x=2

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) 3^2x = 81

    3^2x = 3⁴

=> 2x = 4

=>   x = 2

a). 5^2x-3 – 2.5² = 5² .3 

             5^2x: 5³ = 5² .3 + 2.5² 

             5^2x: 5³ = 5² .5

                   5^2x = 5² .5.5³ 

                   5^2x = 5⁶

               => 2x = 6

                  => x=3

cho B(x) = 0

\(=>-5x+30=0\Rightarrow-5x=-30\Rightarrow x=6\)

cho E(x) = 0

\(=>x^2-81=0\Rightarrow x^2=81=>\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

cho C(x) = 0

\(=>2x+\dfrac{1}{3}=0=>2x=-\dfrac{1}{3}=>x=-\dfrac{1}{6}\)

Cho F(x) = 0

\(=>\left(x-1\right)^2+9=0=>\left(x-1\right)^2=-9\) ( vô lí )

vậy F(x) vô nghiệm

cho D(x) = 0

\(=>\left(x-3\right)\left(16-4x\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\\16-4x=0\end{matrix}\right.\)

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

cho G(x) =0

\(=>\left(x-4\right)\left(x^2+1\right)=0\)

\(=>\left[{}\begin{matrix}x-4=0\\x^2+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\left(tm\right)\\x^2=-1\left(vl\right)\end{matrix}\right.\)

vậy G(x ) có nghiệm là 4

bạn tham khảo hai câu này  nha vì mình ko biết là mấy câu còn lại

B(x)=-5x+30

cho B(x)=0

=> -5x+30=0

-5x=-30

x=-30:(-5)

x=-6

* Vậy nghiệm của đa thức B(x) là -6.

C(x)=2x+1/3

cho C(x)=0

=>2x+1/3=0

2x=-1/3

x=-1/3:2

x=-1/6

vậy nghiệm của đa thức C(x) là -1/6.

`a)(x-6)^2-(x+6)^2=12`

`<=>(x-6-x-6)(x-6+x+6)=12`

`<=>-12.2x=12`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2`

`b)36x^2-12x+1=81`

`<=>(6x-1)^2=81`

`<=>(6x-1-9)(6x-1+9)=0`

`<=>(6x-10)(6x+8)=0`

`<=>(3x-5)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac53\\x=-\dfrac43\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2-6x+2x-12=0`

`<=>x(x-6)+2(x-6)=0`

`<=>(x-6)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2-6x+x-6=0`

`<=>x(x-6)+x-6=0`

`<=>(x-6)(x+1)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

\(a,\\ \left(x+\dfrac{1}{2}\right)^3=\dfrac{8}{125}=\dfrac{2^3}{5^3}\\ \left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow\left(x+\dfrac{1}{2}\right)=\dfrac{2}{5}\\ x=\dfrac{2}{5}-\dfrac{1}{2}\\ x=-\dfrac{1}{10}\)

\(b,3\left|x\right|-27=\dfrac{1}{5}\\ 3\left|x\right|=\dfrac{1}{5}+27\\ 3\left|x\right|=\dfrac{136}{5}\\ \left|x\right|=\dfrac{136}{5}:3\\ \left|x\right|=\dfrac{136}{15}\\ Vậy:x=\dfrac{136}{15}.or.x=-\dfrac{136}{15}\)

\(c,\\ \dfrac{1}{2}.\left(x+\dfrac{1}{3}\right)=\dfrac{1}{16}\\ \left(x+\dfrac{1}{3}\right)=\dfrac{1}{16}:\dfrac{1}{2}\\ \left(x+\dfrac{1}{3}\right)=\dfrac{1}{8}\\ x=\dfrac{1}{8}-\dfrac{1}{3}\\ x=-\dfrac{5}{24}\)

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2