Respuesta :
a.
[tex]\displaystyle\frac1a-\frac1h=\frac1h-\frac1b[/tex]
[tex]\implies\displaystyle\frac{h-a}{ah}=\frac{b-h}{bh}[/tex]
[tex]\implies\displaystyle\frac{h-a}{b-h}=\frac{ah}{bh}[/tex]
[tex]\implies\displaystyle\frac{h-a}{b-h}=\frac ab[/tex]
b.
[tex]\displaystyle h=\frac{2ab}{a+b}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{\frac{2ab}{a+b}-a}{b-\frac{2ab}{a+b}}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{2ab-a(a+b)}{b(a+b)-2ab}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{ab-a^2}{b^2-ab}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{a(b-a)}{b(b-a)}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac ab[/tex]
The other direction can be proved by following the manipulations in the reverse order.
[tex]\displaystyle\frac1a-\frac1h=\frac1h-\frac1b[/tex]
[tex]\implies\displaystyle\frac{h-a}{ah}=\frac{b-h}{bh}[/tex]
[tex]\implies\displaystyle\frac{h-a}{b-h}=\frac{ah}{bh}[/tex]
[tex]\implies\displaystyle\frac{h-a}{b-h}=\frac ab[/tex]
b.
[tex]\displaystyle h=\frac{2ab}{a+b}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{\frac{2ab}{a+b}-a}{b-\frac{2ab}{a+b}}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{2ab-a(a+b)}{b(a+b)-2ab}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{ab-a^2}{b^2-ab}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac{a(b-a)}{b(b-a)}[/tex]
[tex]\displaystyle\implies\frac{h-a}{b-h}=\frac ab[/tex]
The other direction can be proved by following the manipulations in the reverse order.