2^x + 2^x+4 = 272

=> 2^x(1 + 2⁴ ) = 272

=> 2^x . 17 = 272

=> 2^x = 272 : 17 = 16

=> 2^x  = 2⁴

=> x = 4

2^x.17 = 272

2^x = 2⁴

x = 4           

\(2^x+2^{x+4}=272\)

\(< =>2^x.\left(1+2^4\right)=272\)

\(< =>2^x.17=272\)

\(< =>2^x=272:17\)

\(< =>2^x=16\)

\(< =>2^x=2^4\)

\(=>x=4\)

a, x-4/2-15-1/2015=10-2x

qua điều trên:

Ta thấy rằng :

x-5=10-2x

=>

x=15-2x

=>

0=15-3x

=>x=5

Ta có:

2^x+2^x+4=272

=> 2^x.(1+16)=272

=>2^x.17=272

=>2^x=16

=>x=4

a) \(2^{3x+2}=4^{x+5}\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\Leftrightarrow2^{3x+2}=2^{2x+10}\)

\(\Rightarrow3x+2=2x+10\Leftrightarrow3x+2-2x-10\)

\(\Leftrightarrow x-8=0\Leftrightarrow x=8\) vậy \(x=8\)

câu : b ; c mk cảm thấy đề sai

a

\(5\frac{4}{7}:x+=13\)

\(\frac{39}{7}:x=13\)

\(x=\frac{39}{7}:13\)

\(x=\frac{3}{7}\)

\(\frac{4}{7}x=\frac{9}{8}-0,125\)

\(\frac{4}{7}x=1\)

\(x=1:\frac{4}{7}\)

\(x=\frac{7}{4}=1\frac{3}{4}\)

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}