Question

Resolva a inequação :

$\sf{ \sqrt{x} + \sqrt{ \dfrac{1}{x} } \leq 2 }$​

Answer 1

Explicação passo-a-passo:

$\sf \sqrt{x}+\sqrt{\dfrac{1}{x}}\le2$

$\sf \left(\sqrt{x}+\sqrt{\dfrac{1}{x}}\right)^2\le2^2$

$\sf (\sqrt{x})^2+2\cdot\sqrt{x}\cdot\sqrt{\dfrac{1}{x}}+\left(\sqrt{\dfrac{1}{x}}\right)^2\le4$

$\sf x+2+\dfrac{1}{x}\le4$

$\sf x+\dfrac{1}{x}-2\le0$

$\sf x^2-2x+1\le0$

$\sf (x-1)^2\le0$

• $\sf (x-1)^2=0$

$\sf x-1=0$

$\sf x'=x"=1$

Logo, $\sf S=\{1\}$

Answer 2

$\sf{\sqrt{x}+\sqrt{\dfrac{1}{x}}\leq 2}$

Elevando os dois membros ao quadrado temos:

$\sf{(\sqrt{x}+\sqrt{\dfrac{1}{x}}\leq 2^2}\\\sf{x+2\cdot\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}+\dfrac{1}{x}\leq 4}\\\sf{x+2+\dfrac{1}{x}\leq 4}\\\sf{x+2+\dfrac{1}{x}-4\leq 0}\\\sf{x-2+\dfrac{1}{x}\leq 0}\\\sf{x^2-2x+1\leq 0}\\\sf{x^2-2x+1=0}\\\sf{(x-1)^2=0}\\\sf{x-1=0}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{x=1}}}}}$