Bài 1 :

\(C=\frac{1}{\left|x-2\right|+3}\)

\(C\le\frac{1}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy....

Bài 2 :

a) \(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)

\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)

\(\Rightarrow3x-1=5\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b) \(2\cdot3^{x-405}=3^{x-1}\)

\(2=3^{x-1}:3^{x-405}\)

\(2=3^{x-1-x+405}\)

\(2=3^{404}\)( vô lí )

=> x thuộc rỗng

c) \(\frac{1}{81}\cdot27^{2x}=\left(-9\right)^4\)

\(\frac{27^{2x}}{81}=9^4\)

\(\frac{\left(3^3\right)^{2x}}{3^4}=\left(3^2\right)^4\)

\(\frac{3^{6x}}{3^4}=3^8\)

\(3^{6x-4}=3^8\)

\(\Rightarrow6x-4=8\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

d) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}4x-1=0\\4x-1=\left\{\pm1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\left\{\frac{1}{2};0\right\}\end{cases}}\)

a) (3x-7)⁵=32

=> (3x-7)⁵=2⁵

=> 3x-7=2

=> 3x=2+7=9

=>x=9:3=3

b) (4x-1)³=27.125

=> (4x-1)³=3³.5³

=> (4x-1)³=(3.5)³

=> (4x-1)³=15³

=> 4x-1=15

(Các bước còn lại tương tự câu a)

a)(3x-7)^5=32

(3x-7)^5=2^5

3x-7=2

3x=2+7

3x=9

x=9:3

x=3

2^x+2^2x+1=96

2^x+2^2x.2¹=96

2^x+2^2x.2=96

2^x+2^2x=96:2

2^x+2^2x=48

a/ ( x - 1 )⁴ = 81

( x - 1)⁴ = 3⁴

=> x - 1 = 3

     x      = 3 + 1

     x      = 4

b/ ( 3x - 2 )² = 1

( 3x - 2)² = 1²

=> 3x - 2 = 1

     3x      = 1 + 2

     3x      = 3

       x      = 3 : 3

       x      = 1

c/ ( x - 1 )⁵ = -32

( x - 1 )⁵ = (-2)⁵

=> x - 1 = -2

     x      = -2 + 1

     x      = -1

d/ ( 2x - 3 )³ = 125

( 2x - 3 )³ = 5³

=> 2x - 3 = 5

     2x      = 5 + 3

     2x      = 8

       x      = 8 : 2

       x      = 4

k nha bn !!!!

\(\left(x-1\right)^4=81\)

\(\Rightarrow\left(x-1\right)=3\)

\(\Rightarrow x=4\)

\(\left(3x-2\right)^2=1\)

\(\Rightarrow\orbr{\begin{cases}3x-2=-1\\3x-2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

a) (x-1)⁴=3⁴

x-1 =3

x=3+1=4

b) (3x-2)²=1²

3x-2=1

3x=1+2

3x=3

x=3:3

x=1

c)(x-1)⁵=-2⁵

x-1=-2

x=-2+1

x=-1

(2x-3)³=5³

2x-3=5

2x=5+3

2x=8

x=8:2

x=4

a: 3x=81

nên x=27

b: \(5\cdot4^x=80\)

\(\Leftrightarrow4^x=16\)

hay x=2

c: \(2^x=4^5:4^3\)

\(\Leftrightarrow2^x=2^4\)

hay x=4

\(\left(2600+6400\right)-3\cdot x=2000\)

\(9000-3x=2000\)

\(3x=9000-2000\)

\(3x=7000\)

\(x=7000:3\)

\(x=\frac{7000}{3}\)

a) (2x-1)³ = 27
(2x-1)³ = 9³
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2

b) (2x-1)⁴ = 81
(2x-1)⁴ = (\(\pm\)3⁴)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé

À phần b) bạn sửa dòng (2x-1)⁴ = (\(\pm\)3⁴) thành (2x-1)⁴ = (\(\pm\)3)⁴ nhé
Mình vừa viết nhầm

a) ( 3x - 2 )⁵ = -32

<=> ( 3x - 2 )⁵ = -2⁵

<=> 3x - 2 = -2

<=> 3x = 0

<=> x = 0

b) ( 3 - 2x )⁴ = 81

<=> ( 3 - 2x ) = 3⁴

<=> 3 - 2x = 3

<=> 2x = 0

<=> x = 0

c) ( x - 3 )² = ( 3x + 4 )²

<=> ( x - 3 )² - ( 3x + 4 )² = 0

<=> [ x - 3 - ( 3x + 4 ) ][ x - 3 + ( 3x + 4 ] = 0

<=> [ x - 3 - 3x - 4 ][ x - 3 + 3x + 4 ] = 0

<=> [ -2x - 7 ][ 4x + 1 ] = 0

<=> \(\orbr{\begin{cases}-2x-7=0\\4x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)

d) Mời các cao nhân chứ em tịt rồi ạ :((

Sửa đề d nhé : ( x + 1 )^5  thì to quá em sửa thành ( x + 1 ) vì thấy trên mạng cho vậy 

\(\left(2x+5\right)=\left(x+1\right)\Leftrightarrow2x-x+5-1=0\)

\(\Leftrightarrow x=-4\)

a/ \(\left(\frac{1}{5}\right)^x=\left(\frac{1}{5^3}\right)^3=\left(\frac{1}{5}\right)^9\Rightarrow x=9\)

b/ \(\left(\frac{3}{5}\right)^x=\left(\frac{3^2}{5^2}\right)^3=\left(\frac{3}{5}\right)^6\Rightarrow x=6\)

c\(2^{3-2x}=\left(2^3\right)^3=2^9\Rightarrow3-2x=9\Rightarrow x=-3\)

d/ \(2^{3x+1}=32^2=\left(2^5\right)^2=2^{10}\Rightarrow3x+1=10\Rightarrow x=3\)

e/ \(3^{6-3x}=81^3=\left(3^4\right)^3=3^{12}\Rightarrow6-3x=12\Rightarrow x=-2\)

\(\left(\frac{1}{5}\right)^x=\left(\frac{1}{125}\right)^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left[\left(\frac{1}{5}\right)^3\right]^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left(\frac{1}{5}\right)^9\Leftrightarrow x=9\)

\(\left(\frac{3}{5}\right)^x=\left(\frac{9}{25}\right)^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left[\left(\frac{3}{5}\right)^2\right]^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^6\Leftrightarrow x=6\)

\(2^{3-2x}=8^3\Leftrightarrow2^{3-2x}=\left(2^3\right)^3\Leftrightarrow2^{3-2x}=2^9\Leftrightarrow3-2x=9\)

\(\Leftrightarrow2x=3-9\Leftrightarrow2x=-6\Leftrightarrow x=\left(-6\right):2\Leftrightarrow x=-3\)

Các phép còn lại làm tương tự bn nha !

a)\(\left(\frac{1}{5}\right)^x=\left(\left(\frac{1}{5}\right)^3\right)^3=\left(\frac{1}{5}\right)^9\)

=>x=9

b)\(\left(\frac{3}{5}\right)^x=\left(\left(\frac{3}{5}\right)^2\right)^3=\left(\frac{3}{5}\right)^6\)

=>x=6

c) \(2^{3-2x}=\left(2^3\right)^3=2^9\)

=>3-2x=9 

=>2x=-6

=>x=-3

d)\(2^{3x+1}=\left(2^5\right)^2=2^{10}\)

=>3x+1=10

=>x=3

e)\(3^{6-3x}=\left(3^4\right)^3=3^{12}\)

=>6-3x=12

=>3x=-6

=>x=-2