3^x+1 + 3^x+2 - 2 x 3^x = 270

3^x ( 3 + 3² - 2) = 270

3^x . 10 = 270

3^x = 270 : 10

3^x = 27

3^x = 3³

x = 3

\(3^{x+1}+3^{x+2}-2.3^x=270\)

\(\Leftrightarrow3.3^x+3^2.3^x-2.3^x=270\)

\(\Leftrightarrow\left(3+3^2-2\right).3^x=270\)

\(\Leftrightarrow10.3^x=270\)

\(\Leftrightarrow3^x=27\)

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

\(\Leftrightarrow x=3\)

a.\(2^x-2^4.2^7.32=0\)
\(2^x-2^{16}=0\)
\(=>x=16\)
b.\(3^x+3^{x+2}=270\)
\(3^x+3^x.3^2=270\)
\(3^x.10=270\)
\(3^x=27\)
\(=>x=3\)
 

bạn ơi giải chi tiết đi đừng gộp

x²-3x-270=0 

<=> x^2 - 18x + 15x - 270 = 0

<=> x(x - 18) + 15(x - 18) = 0

<=> (x + 15)(x - 18) = 0

<=> x = - 15 hoặc x = 18

vậy_

kết bạn nhé ^^

\(x^2 -3x-270=0\)

\(\Leftrightarrow\)\(( x^2 -18x) + ( 15x-270)=0\)

\(\Leftrightarrow\)\(x. (x-18)+15. (x-18)=0\)

\(\Leftrightarrow\)\((x-18)(x+15)=0\)

\(\Rightarrow\)\(x =18\)  hoặc \(x=-15\)

\(\Leftrightarrow3^x=27\)

hay x=3

\(\Leftrightarrow3^x+3^x.3^2=270\Leftrightarrow3^x.10=270\\\Leftrightarrow3^x=3^3\Leftrightarrow x=3\)

a) 1245 - 3x = 735

               3x = 1245 - 735  

               3x = 510

                x = 510 : 3

                x = 170

b,14x - 8x + 34 = 94

        14x - 8x   =94-34

         14x - 8x  =60

        x(14-8)     = 60

              x.6    =60

                x     =60 :6

                x     =10

\(3^{x+4}+3^{x+2}=270\)

\(\Leftrightarrow3^x\cdot90=270\)

hay x=1

\(3^{x+2}.3^2+3^{x+2}=270\)

\(3^{x+2}\left(3^2+1\right)=270\)

\(3^{x+2}.10=270\)

\(3^{x+2}=27\)

⇒x+2=3

⇒x=1

đặt 3^x ra ngoài bạn nhé bàn phím mik hỏng rồi ;-;

$\Rightarrow 3^x(1+3+3^2+3^3)=1080$

$\Rightarrow 3^x.40=1080$

$\Rightarrow 3^x=27=3^3$

$\Rightarrow x=3$

b: \(3x^2-2x-1=0\)

=>\(3x^2-3x+x-1=0\)

=>\(\left(x-1\right)\left(3x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a: Bạn ghi lại đề đi bạn

Thu gọn biểu thức:

\(9x\left(x+5\right)-\left(3x+2\right)\left(3x-2\right)\)

\(=9x^2+45x-\left(9x^2-4\right)\)

\(=45x+4\)

Tìm x. biết:

\(\left(3x-2\right)^2=49x^2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=7x\\3x-2=-7x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};\dfrac{1}{5}\right\}\)