Ensino Médio ⇒ Radiciação Tópico resolvido

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Simplificar supondo a>0 e b>0:

[tex3]\frac{b-a}{a+b}.[a^{1/2}.(a^{1/2}-b^{1/2})^{-1}-(\frac{a^{1/2}+b^{1/2}}{b^{1/2}})^{-1]}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{b-a}{a+b}.[a^{1/2}.(a^{1/2}-b^{1/2})^{-1}-(\frac{a^{1/2}+b^{1/2}}{b^{1/2}})^{-1]}[/tex3]

Resposta

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-1

[petras]

[tex3]\mathsf{\frac{b-a}{a+b}\cdot \[\frac{a^{\frac{1}{2}}}{(a^{\frac{1}{2}}-b^{\frac{1}{2}})}-\(\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}+b^{\frac{1}{2}}}\)\] =\\\frac{b-a}{a+b}\cdot \[\frac{a^{\frac{1}{2}}\cdot (a^{\frac{1}{2}}+b^{\frac{1}{2}})- b^{\frac{1}{2}}(a^{\frac{1}{2}}-b^{\frac{1}{2}})}{a-b}\]=\\
\frac{b-a}{a+b}\cdot \[\frac{a+ab^{\frac{1}{2}}+ b-ab^{\frac{1}{2}}}{a-b}\]=\\
\frac{(b-a)}{(a+b)}\cdot \frac{(a+b)}{(a-b)}=\frac{b-a}{a-b}=-\frac{(a-b)}{(a-b)}=\boxed{-1}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{\frac{b-a}{a+b}\cdot \[\frac{a^{\frac{1}{2}}}{(a^{\frac{1}{2}}-b^{\frac{1}{2}})}-\(\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}+b^{\frac{1}{2}}}\)\] =\\\frac{b-a}{a+b}\cdot \[\frac{a^{\frac{1}{2}}\cdot (a^{\frac{1}{2}}+b^{\frac{1}{2}})- b^{\frac{1}{2}}(a^{\frac{1}{2}}-b^{\frac{1}{2}})}{a-b}\]=\\
\frac{b-a}{a+b}\cdot \[\frac{a+ab^{\frac{1}{2}}+ b-ab^{\frac{1}{2}}}{a-b}\]=\\
\frac{(b-a)}{(a+b)}\cdot \frac{(a+b)}{(a-b)}=\frac{b-a}{a-b}=-\frac{(a-b)}{(a-b)}=\boxed{-1}}[/tex3]