I.

$\frac{1}{9}=\frac{x+3}{27}$  

$1\cdot 27=9\cdot \left(x+3\right)$  

$27=9\left(x+3\right)\mid :9$  

$3=x+3$  

$x=0$  

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II.      

$\frac{1}{9}=\frac{x-3}{27}$  

$1\cdot 27=9\cdot \left(x-3\right)$  

$27=9\left(x-3\right)\mid :9$  

$3=x-3$  

$x=6$  

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III.

$\frac{4}{3}=\frac{3x+2}{6}$  

$4\cdot 6=3\cdot \left(3x+2\right)$  

$24=9x+6$  

$18=9x\mid :9$  

$x=2$  

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Wpisujemy:

I - C

II - D

III - A