a: \(\Leftrightarrow3^{2x}-10\cdot3^{2x}+81=0\)

\(\Leftrightarrow3^{2x}\cdot\left(-9\right)=-81\)

\(\Leftrightarrow3^{2x}=9\)

=>2x=2

hay x=1

b: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

\(3^{2x}-30.3^x+81=0\\ \Leftrightarrow\left(3^x\right)^2-2.3^x.15+15^2=144\\ \Leftrightarrow\left(3^x-15\right)^2=144\\ \Leftrightarrow\left(3^x-15\right)^2-144=0\\ \Leftrightarrow\left(3^x-15-12\right)\left(3^x-15+12\right)=0\\ \Leftrightarrow\left(3^x-27\right)\left(3^x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3^x-27=0\\3^x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

(x - 2)(2x - 6) = 0

=> x - 2 = 0 hoac 2x - 6 = 0

=> x = 2 hoac x = 3

vay_

(3x + 9)(1 - 3x) = 0

=> 3x + 9 = 0 hoac 1 - 3x = 0

=> x = -3 hoac x = 1/3 

vay_

|2 - x| + 2 = x

=> |2 - x| = x - 2

=> 2 - x = x - 2 hoac 2 - x = 2 - x

=> -2x = -4 hoac x thuoc tap hop rong

=> x = 2 

\(a,\left(x-2\right).\left(2x-6\right)=0\Leftrightarrow2\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(b,\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow3\left(x+3\right).\left(1-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\1-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)

\(c,\left(x^2+1\right).\left(81-x^2\right)=0\Leftrightarrow\left(x^2+1\right).\left(9-x\right).\left(9+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}9-x=0\\9+x=0\end{cases}}\) ( vì \(x^2+1\ne0\forall x\)) \(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)

b . 49x^2 - 81 = 0 

=> 49x^2 = 81 

rồi giải ra nhé 

a ) 2x ( x - 5 ) - x ( 3 + 2x ) = 26

2x² - 10x - 3x - 2x² = 26

- 13x = 26

x = 26 : ( -13 )

x = -2

b) 49x² - 81 = 0

( 7x - 9 )( 7x + 9 ) = 0

Th1 :

7x - 9 = 0

7x = 9

x = \(\frac{9}{7}\)

Th2

7x + 9 = 0

7x = -9

x = \(-\frac{9}{7}\)

Vay x = \(\frac{9}{7}\) hoac x = \(-\frac{9}{7}\)

bn này k bao gio tra cong ng khac, ngu j ma lam

a/ 2x(x-5)-x(3+2x)=26

 =>2x²-10x-2x²-3x=26

=>-13x=26

=>x=-2

b/ 49x²-81=0

=>49x²=81

=>x²=\(\frac{49}{81}\)

\(\Rightarrow x^2=\left(-\frac{9}{7}\right)^2\)hoặc \(\left(\frac{9}{7}\right)^2\)

\(\Rightarrow x=\pm\frac{9}{7}\)

2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

- 13x = 26

x = 26 : (- 13)

x = - 2

49x² - 81 = 0

(7x - 9)(7x + 9) = 0

Th1:

7x - 9 = 0

7x = 9

x = 9/7

Th2:

7x + 9 = 0

7x = - 9 

x = - 9/7

Vậy x = 9/7 hoặc x = - 9/7

\(x^3-25x=0\)

\(x\left(x^2-25\right)=0\)

\(x\left(x-5\right)\left(x+5\right)=0\)

\(x=0,x=5,x=-5\)

\(a,x^3-25x=0\)

\(\Leftrightarrow x\left[x^2-25\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=25\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

Vậy : \(x\in\left\{0;\pm5\right\}\)

\(b,\left[x^2-81\right]\left[18-3x\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-81=0\\18-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=81\\3x=18\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm9\\x=6\end{cases}}\)