\(\left|x\right|=\frac{4}{7}\)

\(\Rightarrow x=\frac{4}{7}\)

b,\(\left(2x-3\right)^2=64\)

\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm8\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=8\\2x-3=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{5}{2}\end{cases}}}\)

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

\(\Rightarrow\left(\frac{1}{2}\right)^x=\left(\pm\frac{1}{2}\right)^4\)

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

d,\(3^{x+1}=27\)

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

\(\Leftrightarrow x+1=3\)

\(\Leftrightarrow x=2\)

(-3/4)63x-1=(3/4)^3

3x-1=3+1

3x=3=1

x=4;3

x=4/3

Vậy x=4/3

25556

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

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

a) \(2^{4x+1}-8^{x+2}=0\)\(\Leftrightarrow2^{4x+1}-2^{3\left(x+2\right)}=0\)

\(\Leftrightarrow2^{4x+1}-2^{3x+6}=0\)\(\Leftrightarrow2^{4x+1}=2^{3x+6}\)

\(\Leftrightarrow4x+1=3x+6\)\(\Leftrightarrow4x-3x=6-1\)\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(3^2.9^{2x}=27^{x+3}\)\(\Leftrightarrow3^2.3^{2.2x}=3^{3\left(x+3\right)}\)\(\Leftrightarrow3^2.3^{4x}=3^{3x+9}\)

\(\Leftrightarrow3^{2+4x}=3^{3x+9}\)\(\Leftrightarrow2+4x=3x+9\)\(\Leftrightarrow4x-3x=9-2\)\(\Leftrightarrow x=7\)

Vậy \(x=7\)

c) \(8^{2x}.64^2=16^{x+4}\)\(\Leftrightarrow2^{3.2x}.2^{6.2}=2^{4\left(x+4\right)}\)\(\Leftrightarrow2^{6x}.2^{12}=2^{4\left(x+4\right)}\)

\(\Leftrightarrow2^{6x+12}=2^{4x+16}\)\(\Leftrightarrow6x+12=4x+16\)\(\Leftrightarrow6x-4x=16-12\)

\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

2/

a/ \(25x^2-1=0\)

<=> \(\left(5x\right)^2-1=0\)

<=> \(\left(5x-1\right)\left(5x+1\right)=0\)

<=> \(\orbr{\begin{cases}5x-1=0\\5x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-\frac{1}{5}\end{cases}}\)

b/ \(4\left(x-1\right)^2-9=0\)

<=> \(\left[2\left(x-1\right)\right]^2-3^2=0\)

<=> \(\left(2x-2\right)^2-3^2=0\)

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

<=> \(\left(2x-5\right)\left(2x+1\right)=0\)

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

c/ \(\frac{1}{4}-9\left(x+1\right)^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left[3\left(x-1\right)\right]^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left(3x-3\right)^2=0\)

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

<=> \(\left(\frac{7}{2}-3x\right)\left(-\frac{5}{2}+3x\right)=0\)

<=> \(\orbr{\begin{cases}\frac{7}{2}-3x=0\\-\frac{5}{2}+3x=0\end{cases}}\)<=> \(\orbr{\begin{cases}3x=\frac{7}{2}\\3x=\frac{5}{2}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{5}{6}\end{cases}}\)

d/ \(\frac{1}{16}-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}\right)^2-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}-2x-\frac{3}{4}\right)\left(\frac{1}{4}+2x+\frac{3}{4}\right)=0\)

<=> \(\left(-\frac{1}{2}-2x\right)\left(1+2x\right)=0\)

<=> \(2\left(-\frac{1}{4}-x\right)\left(1+2x\right)=0\)

<=> \(\orbr{\begin{cases}-\frac{1}{4}-x=0\\1+2x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{1}{2}\end{cases}}\)

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

ko bt có đúng ko nữa, mấy câu kia tui ko bt lm

Sos

hahaha

a) Ta có: \(15.3^x-1=1215\)

                          \(3^{x-1}=81=3^4\)

                        \(x-1=4\)

                                  \(x=5\)

Vậy \(x=5\)

b) Ta có: \(\left(2x-1\right).3=27\)

                          \(2x-1=9\)

                                    \(2x=10\)

                                       \(x=5\)

Vậy \(x=5\)

a) 

\(15\cdot3^{x-1}=1215\) 

\(3^{x-1}=1215:15\) 

\(3^{x-1}=81\) 

\(3^{x-1}=3^4\)   

\(x-1=4\) 

\(x=4+1\)  

\(x=5\) 

b) 

\(\left(2x-1\right)\cdot3=27\) 

\(2x-1=27:3\) 

\(2x-1=9\) 

\(2x=9+1\) 

\(2x=10\) 

\(x=10:2\) 

\(x=5\) 

c) 

\(\left(x+1\right)^2=64\) 

\(x+1=\pm\sqrt{64}=\pm8\) 

\(\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}}\) 

\(\orbr{\begin{cases}x=8-1\\x=-8-1\end{cases}}\) 

\(\orbr{\begin{cases}x=7\\x=-9\end{cases}}\)

a) (x – 45).27 = 0 ó x – 45 = 0 ó x = 45

b) 45.(2x – 4).13 = 0 ó 2x – 4 = 0 ó 2x – 4 = 0 ó 2x = 4 ó x = 2

c) (x – 3).(x – 5) = 0

Vậy tập nghiệm của phương trình là S = {3;5}

a x=45+0=45