3^ x -1 = 1/243

3^x =1/243 +1

3^x = 244 / 243

Ta  thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn

81^-2x . 27^x =9^5

81^-2 . 81^x . 27^x =9^5

1/9^4 . (81.27)^x =9 ^5

            3^6x = 9^5 : 1/9^4

             3^6x = 9^9

            3^6x = 3^18

    => 6x =18

  x=3

2^x +2^x +3 =144

2.(2^x) =141

2^x+1 = 141

Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn

a. (x - 2)² = 1

<=> (x - 2)² = 1² = (-1)²

<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)

Vậy x \(\in\){1; 3}.

b. (2x - 1)³ = -8

<=> (2x - 1)³ = (-2)³

<=> 2x - 1 = -2

<=> 2x = -2 + 1

<=> 2x = -1

<=> x = -1/2

Vậy x = -1/2.

c. (x + 1/2)² = 1/16

<=> (x + 1/2)² = (1/4)² = (-1/4)²

<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)

Vậy x \(\in\){-1/4; -3/4}.

d. (x - 2)³ = -27

<=> (x - 2)³ = (-3)³

<=> x - 2 = -3

<=> x = -3 + 2

<=> x = -1

Vậy x = -1.

a.\(\left(x-2\right)^2\)=1

<=> x-2=1 hoặc x-2=-1

<=> x= 3 hoặc x=1

b.\(\left(2x-1\right)^3\)=-8

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

2x-1=-2

2x=-1

x=-1/2

c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)

\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)

x+\(\frac{1}{2}\)=\(\frac{1}{4}\)  hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)

x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)

d.\(\left(x-2\right)^3\)=-27

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

x-2=-3

x=-1

g.\(3^{x-1}\)=\(\frac{1}{243}\)

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

\(3^{x-1}\)=\(3^{-5}\)

x-1=-5

x=-4

a) \(81^{-2x}.27^x=9^5\)

\(\Rightarrow\left(3^4\right)^{-2x}.\left(3^3\right)^x=3^{10}\)

\(\Rightarrow3^{-8x}.3^{3x}=3^{10}\)

\(\Rightarrow3^{-5}=3^{10}\)

\(\Rightarrow-5x=10\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

b) \(2^x+2^{x+3}=144\)

\(\Rightarrow\left(1+2^3\right).2^x=144\)

\(\Rightarrow\left(1+8\right).2^x=144\)

\(\Rightarrow9.2^x=144\)

\(\Rightarrow2^x=16\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

c) \(2^{x-1}+5.2^{x-2}=7.32\)

\(\Rightarrow\left(2+5\right).2^{x-2}=244\)

\(\Rightarrow7.2^{x-2}=244\)

\(\Rightarrow2^{x-2}=32\)

\(\Rightarrow x-2=5\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

a) (X - 2)\(^2\) = 1 <=> X - 2 = \(\sqrt{1}\) <=> X = 1 + 2 <=> X = 3

a) \(\left(2x+3\right)^2=\frac{9}{144}\)

\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x+3=\frac{1}{4}\\2x+3=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{-11}{4}\\2x=\frac{-13}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-11}{8}\\x=\frac{-13}{8}\end{cases}}}\)

Vậy ...

b) Ta có: \(\left(3x-1\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)

Vậy ....

c) \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25\Leftrightarrow x=\left\{5;-5\right\}\)

Vậy ...

d) \(\frac{x^7}{81}=27\Leftrightarrow x^7=27.81=2187\)

Mà 3⁷ = 2187 => x⁷ = 3⁷ => x = 3

Vậy ....

e) \(\frac{x^8}{9}=729\Leftrightarrow x^8=729.9=6561\)

Mà 3⁸ = (-3)⁸ = 6561

=> x⁸ = 3⁸ = (-3)⁸

=> x = {-3;3}

Vậy ...

a)2³ .2^x=128

=>2^3+x=128

=>2^3+x=2⁷

=>3+x=7

=>x=4

b)2^x+2^x+3=144

=>2^x(1+2³)=144

=>2^x*9=144

=>2^x=16

=>2^x=2⁴

=>x=4

c)x-2⁸⁷:2⁸⁵=125

=>x-2^87-85=125

=>x-4=125

=>x=129

d)3^x+4-3^x=720

=>3^x(3⁴-1)=720

=>3^x*80=720

=>3^x=9

=>3^x=3²

=>x=2

e)81^2x:27^x=9⁵

=>(3⁴)^2x:(3³)^x=9⁵

=>3^8x:3^3x=9⁵

=>3^8x-3x=(3²)⁵

=>3^5x=3¹⁰

=>5x=10

=>x=2

a) 2³ .2^x=128

=>2^3+x = 2⁷

=>3+x=7

=>x=4

Bài 1: 

a) \(x^3-16x=x\left(x-4\right)\left(x+4\right)\)

b) \(3x^2+3y^2-6xy-12=3\left(x^2-2xy+y^2-4\right)=3\left(x-y-2\right)\left(x-y+2\right)\)

c) \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)

d) \(x^4+x^3+2x^2+x+1=\left(x^2+x+1\right)\left(x^2+1\right)\)

Bài 2: 

a) Ta có: \(\left(x+6\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=12\\x+6=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-18\end{matrix}\right.\)

b) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)

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

c) Ta có: \(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)

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