a)  1999^{2x - 6} = 1 

1999^{2x - 6} = 1999⁰

=>2x-6=0

2x=0+6

2x=6

x=6:2

x=3

b)  2^x x 4 = 128

2^x=128:4

2^x=32

2^x=2⁵

=>x=5

c)  ( 2x + 1 ) ³ = 125

( 2x + 1 ) ³ = 5³

2x+1=5

2x=5-1

2x=4

x=4:2

x=2

a)1999^2x-6=1

1999^2x-6=1999⁰

=>2x-6=0

2x=0+6

2x=6

x=6:2

x=3

b)2^x.4=128

2x=128:4

2x=32

x=32:2

x=5

c)(2x+1)³=125

(2x+1)³=5³

2x+1=5

2x=5-1

2x=4

x=4:2

x=2

Ta có : (2x + 1)³ = 125

=> (2x + 1)³ = 5³

=> 2x + 1 = 5

=> 2x = 4

=> x = 2

Ta có : (4x - 1)² = 25 x 9

=> (4x - 1)² = 5².3²

=> (4x - 1)² = 15²

\(\Leftrightarrow\orbr{\begin{cases}4x-1=15\\4x-1=-15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=16\\4x=-14\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{7}{2}\end{cases}}\)

\(a,4^{2x-6}=1\)\(\Leftrightarrow2x-6=0\Leftrightarrow2x=6\Rightarrow x=3\)

\(b,2^{2x-1}=16\Rightarrow2^{2x-1}=2^4\)

\(\Rightarrow2x-1=4\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)

\(c,5< 5^x< 125\Rightarrow5^1< 5^x< 5^3\)\(\Rightarrow1< x< 3\)

\(d,5^{x+1}=125\Rightarrow5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2\)

\(a,4^{2x-6}=1\)

\(4^{2x-6}=4^0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\Leftrightarrow x=3\)

\(b,2^{x-1}=16\)

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

\(\Rightarrow x-1=4\)

\(\Rightarrow x=5\)

Nếu bài dễ thì cố gắng tự làm bạn nhé.

Bài 1:

a) \(2x+3=5\)

\(\Leftrightarrow2x=5-3=2\)

\(\Leftrightarrow x=2:2=1\)

Vậy: x=1

b) \(\left(2x-3\right)^2=9\)\(=3^2=\left(-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left(3+3\right):2=3\\x=\left(-3+3\right):2=0\end{matrix}\right.\)

Vậy: \(x\in\left\{3;0\right\}\)

c) \(3x+4=2x-7\)

\(\Leftrightarrow2x-3x=7+4\)

\(\Leftrightarrow-1x=11\)

\(\Leftrightarrow x=-11\)

Vậy: x=-11

d) \(\left(4x-1\right)^3=27\)\(=3^3\)

\(\Rightarrow4x-1=3\)

\(\Leftrightarrow4x=3+1=4\)

\(\Leftrightarrow x=4:4=1\)

Vậy: x=1

e) \(\left(x-7\right).2=3.\left(2x+4\right)\)

\(\Leftrightarrow2x-14=6x+12\)

\(\Leftrightarrow2x-6x=12+14\)

\(\Leftrightarrow-4x=26\)

\(\Leftrightarrow x=\frac{26}{-4}=\frac{13}{-2}\)

Vậy: \(x=\frac{13}{-2}\)

A)                    \(\left(2x+1\right)^2=9^3\)

                    \(\Rightarrow\left(2x+1\right)^2=9^2\times9\)

                    \(\Rightarrow2x+1=81\)

                    \(\Rightarrow2x=81-1\)

                    \(\Rightarrow2x=80\)

                    \(\Leftrightarrow x=40\)

B)                 \(\left(2x+1\right)^3=125^2\)

                      \(\Rightarrow\left(2x+1\right)^3=5^3\)

                      \(\Rightarrow2x+1=5\)

                      \(\Rightarrow2x=5-1\)

                      \(\Leftrightarrow x=2\)

C)                   \(25^2< 5^{x+3}< 25^4\)

                      \(\Leftrightarrow5^4< 5^{x+3}< 5^6\)

                      \(\Leftrightarrow4< x+3< 6\)

                      \(\Rightarrow x+3=5\)

                      \(\Rightarrow x=2\)

D)                   \(x^{15}=x\)

Nếu \(x>1\)thì \(x^{15}>x\)

Vậy \(x=1\)

E)                   \(\left(x-5\right)^4=\left(x-5\right)^6\)

                      \(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)

                      \(\Rightarrow\left(x-5\right)\times\left(1^4-1^6\right)=0\)

                     \(\Rightarrow x-5=0\)

                       \(\Leftrightarrow x=5\)

KÍCH MK NHA BẠN

a) x=5

b) x=0 hoặc 1

c) x=2

d) x=6 hoặc 5

e) x=0 hoặc 1

f) x=8

a, 2^x . 4 = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

=> x = 5

b, x¹⁵ = x¹

=> x¹⁵ - x = 0

x . ( x¹⁴ - 1 ) = 0

=> x = 0 hoặc x¹⁴ - 1 = 0 

=> x = 0 hoặc x = 1

c, (2x + 1)³ = 125

( 2x + 1 )³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

d, (x – 5)⁴ = (x - 5)⁶

=> ( x - 5 )⁶ - ( x - 5 )⁴ = 0

=> ( x - 5 )⁴ . [ ( x - 5 )² - 1 ] = 0

=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

e, x¹⁰ = x

x¹⁰ - x = 0

x . ( x⁹ - 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

f, (2x -15)⁵ = (2x -15)³

( 2x - 15 )⁵ - ( 2x - 15 )³ = 0

( 2x - 15 )³ . [ ( 2x - 15 )² - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}x\text{ không tồn tại}\\x=8\end{cases}}}\)

ko biet

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