1)
Folosim formula fundamentală a proporțiilor, sau folosim cazul în care un raport este obținut din celălalt, prin amplificare .
[tex]\it \dfrac{2}{7} = \dfrac{6}{21}\ \ (al\ doilea\ raport\ se\ ob\c{\it t}ine\ din\ primul,\ prin\ amplificare\ cu\ 3)\\ \\ \\ \dfrac{225}{900} = \dfrac{1}{4}\ \ (primul\ raport\ se\ ob\c{\it t}ine\ din\ al\ doilea,\ prin\ amplificare\ cu\ 225)\\ \\ \\ \dfrac{3}{7}\ne \dfrac{18}{420}, \ corect \ ar\ fi \ \dfrac{3}{7} =\dfrac{18}{42},\ (amplificarea\ cu\ 6\ este\ vizibil\breve{a} )[/tex]
[tex]\it \dfrac{1,2}{3} \ne\dfrac{1}{5},\ deoarece\ \begin{cases} \it 1,2\cdot5=6\\ \\ \it 3\cdot1=3\end{cases}\\ \\ \\ \dfrac{1,2}{3} = \dfrac{2}{5}\ (produsul\ extremilor\ este\ egal\ cu produsul\ mezilor)[/tex]
2)
[tex]\it \dfrac{\ 99^{(3}}{165} = \dfrac{\ 33^{(11}}{55}=\dfrac{3}{5}\\ \\ \\ \dfrac{\ 8^{(4}}{3,2} =\dfrac{2}{0,8}\\ \\ \\ \dfrac{\ 4,2^{(2}}{1,4} = \dfrac{2,1}{0,7}[/tex]
3)
[tex]\it \dfrac{^{10)}5}{\ \ 7} = \dfrac{50}{70};\ \ \ \dfrac{^{15)}5}{\ \ 7} = \dfrac{75}{105}\\ \\ \\ \dfrac{^{10)}6}{\ \ 11} = \dfrac{60}{110};\ \ \dfrac{^{15)}6}{\ \ 11} = \dfrac{90}{165}\\ \\ \\ \dfrac{^{10)}0,5}{\ \ 0,3} = \dfrac{5}{3};\ \ \ \dfrac{^{15)}0,5}{\ \ 0,3} = \dfrac{7,5}{4,5}[/tex]
4)
[tex]\it\dfrac{^{100)}1,98}{\ \ 49,5} =\dfrac{198^{(9}}{4950}=\dfrac{\ 22^{(11}}{550} =\dfrac{\ 2^{(2}}{50} =\dfrac{1}{25}\\ \\ \\ Da,\ scara\ este\ \ \dfrac{1}{25}.[/tex]
