a)    9^x-1=3²

( 3² )^x-1 = 3²

 3^2x-2    = 3²

⇒ 2x-2 = 2

    2x    = 2+2

    2x    = 4

   x       = 4 : 2

   x       = 2

b) 5^x+2=625

    5^x+2= 5⁴

⇒ x+2 = 4

    x     = 4-2

   x      = 2

c) 2^x: 2⁵= 2

    2^x:2⁵ = 2¹

    2^x      = 2¹ . 2⁵

    2^x      = 2⁶

⇒ x        = 6

d) 3^x:27=3

    3^x:3³ = 3¹

     3^x     = 3¹.3³

    3^x     = 3⁴

⇒ x      = 4

a) Ta có: \(9^{x-1}=3^2\)

\(\Leftrightarrow3^{2x-2}=3^2\)

\(\Leftrightarrow2x-2=2\)

\(\Leftrightarrow2x=4\)

hay x=2

Vậy: x=2

b) Ta có: \(5^{x+2}=625\)

\(\Leftrightarrow5^{x+2}=5^4\)

\(\Leftrightarrow x+2=4\)

hay x=2

Vậy: x=2

c) Ta có: \(2^x:2^5=2\)

\(\Leftrightarrow2^{x-5}=2^1\)

\(\Leftrightarrow x-5=1\)

hay x=6

Vậy: x=6

d) Ta có: \(3^x:27=3\)

\(\Leftrightarrow3^x:3^3=3\)

\(\Leftrightarrow3^{x-3}=3^1\)

\(\Leftrightarrow x-3=1\)

hay x=4

Vậy: x=4

cần kết quả đúng ko 

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)

a) ( x+ 3 ) ( x - 3 ) = 3 ( x-3)

x+ 3 =3

x =0

a) x² - 9 = 3( x - 3 )

⇔ ( x - 3 )( x + 3 ) - 3( x - 3 ) = 0

⇔ ( x - 3 )( x + 3 - 3 ) = 0

⇔ ( x - 3 ).x = 0

⇔ x - 3 = 0 hoặc x = 0

⇔ x = 3 hoặc x = 0

b) 3( 3x² + 1 ) = 6 - 2( 3x + 2 )

⇔ 9x² + 3 = 6 - 6x - 4

⇔ 9x² + 6x + 3 - 6 + 4 = 0

⇔ 9x² + 6x + 1 = 0

⇔ ( 3x + 1 )² = 0

⇔ 3x + 1 = 0

⇔ x = -1/3

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)