a) 3^3x+1 . 5 = 10935

=> 3^3x . 3 . 5 = 10935

=> 27^x . 3 . 5 = 10935

=> 27^x = 729

=> 27^x = 27²

=> x = 2

b) 2^{5x + 1} . 3 = 6144

=> 2^5x . 2 . 3 = 6144

=> 32^x . 6 = 6144

=> 32^x = 1024

=> 32^x = 32²

=> x = 2

c) (2x + 3)⁴ = 625

=> (2x + 3)⁴ = (\(\pm\)5)⁴

=> \(\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

d) (3x + 1)³ = 343

=> (3x + 1)³ = 7³

=> 3x + 1 = 7 => 3x = 6 => x = 2

a) 3^{3x + 1}. 5 = 10935

3^3x+1 = 10935 : 5

3^3x+1 = 2187

3^3x+1 = 3⁷

=> 3x + 1 = 7

      3x = 7 - 1

      3x = 6

        x = 6:3

   => x = 2

a) \(3^{3x+1}\cdot5=10935\)

\(\Leftrightarrow3^{3x+1}=2187=3^7\)

\(\Rightarrow3x+1=7\)

\(\Leftrightarrow3x=6\)

\(\Rightarrow x=2\)

b) \(2^{5x+1}\cdot3=6144\)

\(\Leftrightarrow2^{5x+1}=2048=2^{11}\)

\(\Leftrightarrow5x+1=11\)

\(\Leftrightarrow5x=10\)

\(\Rightarrow x=2\)

a: =>2x+5=4

=>2x=-1

hay x=-1/2

b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)

=>(3x-4)(3x-5)(3x-3)=0

hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)

c: \(\Leftrightarrow3^{x+1}=3^{2x}\)

=>2x=x+1

=>x=1

d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)

=>2x+3=2x-10

=>0x=-13(vô lý)

( 3/7 ) 5 x ( 7/3 ) - 1 x ( 5/3 )6 : ( 343/652) - 2

= ( 3/7 x 7/3 ) + 5 - 1 x ( 5/3 : 343/652 ) - 2

= 1 + 5 - 1 x 2 - 1715/1956

= 6 - 2 - 1715/1956

=4 - 1715/1956

= 6109/1715

bn ê, đề là j vậy,sao mk đọc mà ko hỉu

ban ghi gi the (3/7)5 nghia la gi

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

                \(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

                                    \(2x=\frac{3}{5}-\frac{3}{5}\)

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

\(\left(3x-1\right)\left(-\frac{x}{2}+5\right)=0\)

\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\5-\frac{x}{2}=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

a) 2x + 23x = 625

   x (2 + 23) = 625

   x. 25        = 625

             x    = 625 : 25

             x    = 25

b) (2^2+4^2) x +2^4 x 5 * x = 100 (dấu* là dấu nhân vì mik không muốn trùng nhau nên viết vậy)

    ( 4 + 16 ) x + 2^4 x 5 * x = 100

         20 * x +   16 x 5 * x =100

            x ( 20 + 16 x 5 )  = 100

           x ( 20 + 80 ) = 100

           x * 100 =100

                   x=100 : 100

                   x= 1

c) ( 3x +1)^2 = 6^2+8^2

    (3x +1)^2  = 36 +64

    (3x +1)^2   =100

     (3x +1)^2  = 10^2

           3x + 1 =10

                  3x=10 - 1=9

                    x = 9 :3

                    x = 3

( d và e làm tương tự )

k mik nha!

Dể mà,k mik mik giải cho

Giai giup minh di ban

\(a)\left(2x+1\right)^2=25\)

\(\Rightarrow\left(2x+1\right)^2=\left(\pm5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

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

\(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\frac{9}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

\(b)5^x+2=625\)

\(\Rightarrow5^x=623\)

\(\Rightarrow x\in\varnothing\)

Vậy \(x\in\varnothing\)

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

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

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-1\)

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

\(c)\left(x-3\right)^2+\left(15x-45\right)^4=0\)

- Có \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0,\forall x\\\left(15x-45\right)^4\ge0,\forall x\end{matrix}\right.\Rightarrow\left(x-3\right)^2+\left(15x-45\right)^4\ge0,\forall x\)

Suy ra: Để \(\left(x-3\right)^2+\left(15x-45\right)^4=0\) thì \(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(15x-45\right)^4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-3=0\\15x-45=0\end{matrix}\right.\Rightarrow x=3\)

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

- Có \(\left\{{}\begin{matrix}\left|x-3\right|\ge0,\forall x\\\left(x^2-3x\right)^2\ge0,\forall x\end{matrix}\right.\Rightarrow\left|x-3\right|+\left(x^2-3x\right)^2\ge0,\forall x\)

Suy ra: Để \(\left|x-3\right|+\left(x^2-3x\right)^2=0\)thì \(\left\{{}\begin{matrix}\left|x-3\right|=0\\\left(x^2-3x\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-3=0\\x^2-3x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x\left(x-3\right)=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)