Zadatak 0226

Test: MKsIIJAII_046

 

Reši po x jednačinu (a i b su realne konstante):

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

 

Rešenje:


 

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

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

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

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

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

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

 

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

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

\(2a^2+2ab-2ax-2bx-3ax=0\)

\(x(-5a-2b)=-2a^2-2ab\)

\(x=\frac {2a(a+b)}{5a+2b}\)

 

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

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

\(2a^2+2ab-2ax-2bx+3ax=0\)

\(x(a-2b)=-2a^2-2ab\)

\(x=\frac {2a(a+b)}{a-2b}\).