Question

Ex 17, va rog mult!! Dau coroana!!

Answer 1

   
[tex]\displaystyle\\ \frac{1}{2} \cdot \left\{\frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot \left( \frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]-\frac{1}{2^{12}} \right\}+\frac{1}{2^{13}}=\frac{1}{2^{12}}\\\\\\ \frac{1}{2} \cdot \left\{\frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot \left( \frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]-\frac{1}{2^{12}} \right\}=\frac{1}{2^{12}}-\frac{1}{2^{13}}\\\\\\ [/tex]


[tex]\displaystyle\\ \frac{1}{2} \cdot \left\{\frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot \left( \frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]-\frac{1}{2^{12}} \right\}=\frac{2}{2^{13}}-\frac{1}{2^{13}}\\\\\\ \frac{1}{2} \cdot \left\{\frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot \left( \frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]-\frac{1}{2^{12}} \right\}=\frac{1}{2^{13}} [/tex]


[tex]\displaystyle\\ \frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot \left( \frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]-\frac{1}{2^{12}} =\frac{1}{2^{13}}:\frac{1}{2}\\\\\\ \frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot \left( \frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]-\frac{1}{2^{12}} =\frac{1}{2^{12}}\\\\\\ \frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot \left( \frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]=\frac{1}{2^{12}}+\frac{1}{2^{12}} [/tex]


[tex]\displaystyle\\ \frac{1}{2^2}\cdot\left[\frac{1}{2^3}\cdot\left(\frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}\right]=\frac{1}{2^{11}}\\\\\\ \frac{1}{2^3}\cdot\left(\frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}=\frac{1}{2^{11}}:\frac{1}{2^2}\\\\\\ \frac{1}{2^3}\cdot \left(\frac{1}{2^4}-\frac{x}{2^5}\right)-\frac{1}{2^9}=\frac{1}{2^{9}}\\\\\\ \frac{1}{2^3}\cdot\left(\frac{1}{2^4}-\frac{x}{2^5}\right)=\frac{1}{2^{9}}+\frac{1}{2^9} [/tex]


[tex]\displaystyle\\ \frac{1}{2^3}\cdot\left(\frac{1}{2^4}-\frac{x}{2^5}\right)=\frac{1}{2^8}\\\\\\ \frac{1}{2^4}-\frac{x}{2^5}=\frac{1}{2^8}:\frac{1}{2^3}\\\\\\ \frac{1}{2^4}-\frac{x}{2^5}=\frac{1}{2^5}\\\\\\ \frac{1}{2^4}-\frac{x}{2^5}=\frac{1}{2^5}\\\\\\ \frac{x}{2^5}=\frac{1}{2^4}-\frac{1}{2^5}\\\\\\ \frac{x}{2^5}=\frac{1}{2^5}\\\\ \boxed{\bf x = 1}[/tex]