Question

calcula o domínio de existência de cada expressão:
a)√(x+1)²
b)⁴√x-1
------
x+1

c)√x-5
--------
x-5
2Racionalize cada uma das seguintes expressões
a)x+2
-------
√x + √2

b) 3
--------
2✓5

c) 1
-------
√2+1

3-resolva as equações irracionais
a).
|------
√3x+5= -2
b)-------. ---------
√3x+1- √x-1=0​

Answer 1

$\large\boxed{\begin{array}{l}\rm 1)\\\tt a)~\sf\sqrt{(x+1)^2}\\\sf (x+1)^2\geqslant0,\forall~x\in\mathbb{R}\\\sf portanto~Dom\,f(x)=\{x\in\mathbb{R}\}\\\tt b)~\sf\sqrt{x-1}\\\sf x-1\geqslant0\\\sf x\geqslant1\\\sf Dom~f(x)=\{x\in\mathbb{R}/x\geqslant1\}\\\tt c)~\sf\dfrac{\sqrt{x-5}}{x-5}\\\\\sf x-5>0\\\sf x>5\\\sf Dom~f(x)=\{x\in\mathbb{R}/x>5\}\end{array}}$

$\boxed{\begin{array}{l}\rm 2)\\\tt a)~\sf\dfrac{(x+2)}{(\sqrt{x}+\sqrt{2})}\cdot\dfrac{(\sqrt{x}-\sqrt{2})}{(\sqrt{x}-\sqrt{2})}\\\\\sf\dfrac{(x+2)}{(\sqrt{x})^2-(\sqrt{2})^2}=\dfrac{x+2}{x-2}\\\\\tt b)~\sf\dfrac{3}{2\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{3\sqrt{5}}{2\cdot(\sqrt{5})^2}=\dfrac{3\sqrt{5}}{2\cdot5}\\\\\sf=\dfrac{3\sqrt{5}}{10}\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf\dfrac{1}{(\sqrt{2}+1)}\cdot\dfrac{(\sqrt{2}-1)}{(\sqrt{2}-1)}=\dfrac{\sqrt{2}-1}{(\sqrt{2})^2-1^2}\\\\\sf\dfrac{\sqrt{2}-1}{2-1}=\dfrac{\sqrt{2}-1}{1}=\sqrt{2}-1\end{array}}$

$\boxed{\begin{array}{l}\rm 3)\\\tt a)~\sf\sqrt{3x+5}=-2\\\sf essa~equac_{\!\!,}\tilde ao~n\tilde ao~faz~o~menor~sentido~em~\mathbb{R}\\\sf pois~neste~conjunto~toda~raiz~quadrada~\acute e~positiva.\end{array}}$

$\boxed{\begin{array}{l}\tt b)~\sf\sqrt{3x+1}-\sqrt{x-1}=0\\\sf \sqrt{3x+1}=\sqrt{x-1}\\\sf(\sqrt{3x+1})^2=(\sqrt{x-1})^2\\\sf 3x+1=x-1\\\sf 3x-x=-1-1\\\sf 2x=-2\\\sf x=-\dfrac{2}{2}\\\sf x=-1\\\underline{\rm verificac_{\!\!,}\tilde ao\!:}\\\sf \sqrt{3\cdot(-1)+1}=\sqrt{-3+1}=\sqrt{-2}\not\in\mathbb{R}\\\sf \sqrt{-1-1}=\sqrt{-2}\not\in\mathbb{R}\\\sf S=\bigg\{~~\bigg\}\end{array}}$