Zadatak 0285

Test: MKsIIJAIII_059

 

Reši jednačinu po x (a, b su parametri):

 

\(\frac{\left ( \frac{a-x}{x} \right )^2-\left ( \frac{a}{a+b} \right )^2}{x^2-2ax+a^2}=\frac{5}{9x^2}\).

 

Rešenje:


 

 

\(...\Leftrightarrow \frac{\left ( \frac{a-x}{x} \right )^2-\left ( \frac{a}{a+b} \right )^2}{(a-x)^2}=\frac{5}{9x^2}\)    \(\Leftrightarrow \left ( \frac{a-x}{x} \right )^2-\left ( \frac{a}{a+b} \right )^2=\frac{5(a-x)^2}{9x^2}\)

 

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

 

\(\Leftrightarrow \frac{4(a-x)^2}{9x^2}-\left ( \frac{a}{a+b} \right )^2=0\)

 

\(\Leftrightarrow \left (\frac{2}{3}\cdot \frac{a-x}{x} -\frac{a}{a+b} \right )\left (\frac{2}{3}\cdot \frac{a-x}{x} +\frac{a}{a+b} \right )=0\) 

 

\(\Leftrightarrow \frac{2}{3}\cdot \frac{a-x}{x} -\frac{a}{a+b}  =0\vee \frac{2}{3}\cdot \frac{a-x}{x} +\frac{a}{a+b} =0\)

 

\(\Leftrightarrow \frac{2}{3}\cdot \frac{a-x}{x} =\frac{a}{a+b} \vee \frac{2}{3}\cdot \frac{a-x}{x} =-\frac{a}{a+b} \)

 

\(\Leftrightarrow (2a-2x)(a+b)=3ax\vee (2a-2x)(a+b)=-3ax\)  

 

\(\Leftrightarrow 2a(a+b)-2(a+b)x-3ax=0 \vee 2a(a+b)-2(a+b)x+3ax=0\)

 

\(\Leftrightarrow x(-2a-2b-3a)=-2a(a+b) \vee x(-2a-2b+3a)=-2a(a+b)\) 

 

\(\Leftrightarrow x_{1}=\frac{2a(a+b)}{5a+2b} \vee x_{2}=-\frac{2a(a+b)}{a-2b}\).