Matemática, perguntado por alic157, 11 meses atrás

Racionalize o denominador das seguintes expressões fracionária:​

Anexos:

Soluções para a tarefa

Respondido por CyberKirito
4

1)

a)

\mathsf{\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}}

b)

\mathsf{\dfrac{2}{\sqrt{10}}=\dfrac{2\sqrt{10}}{10}=\dfrac{\sqrt{10}}{5}}

c)

\mathsf{\dfrac{6}{\sqrt{2}}=\dfrac{6\sqrt{2}}{2}=3\sqrt{2}}

d)

\mathsf{\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}}

e)

\mathsf{\dfrac{6}{5\sqrt{3}}=\dfrac{6\sqrt{3}}{5</p><p>.3}=\dfrac{2\sqrt{3}}{5}}

f)

\mathsf{\dfrac{1}{3\sqrt{2}}=\dfrac{\sqrt{2}}{6}}

g)

\mathsf{\dfrac{a}{\sqrt{a}}=\dfrac{a\sqrt{a}}{a}=\sqrt{a}}

h)

\mathsf{\dfrac{5\sqrt{2}}{2\sqrt{5}}=\dfrac{5\sqrt{10}}{2.5}=\dfrac{\sqrt{10}}{2}}

i)

\mathsf{\dfrac{ab}{\sqrt{b}}=\dfrac{ab\sqrt{b}}{b}=a\sqrt{b}}

j)

\mathsf{\dfrac{R}{2\sqrt{R}}=\dfrac{R\sqrt{R}}{2R}=\dfrac{\sqrt{R}}{2}}

O professor almoçou a letra k portanto vamos para a L.

l)

\mathsf{\dfrac{\sqrt{x}}{2\sqrt{y}}=\dfrac{\sqrt{xy}}{2y}}

m)

\mathsf{\dfrac{2}{a\sqrt{2}}=\dfrac{2\sqrt{2}}{2a}=\dfrac{\sqrt{2}}{a}}

n)

\mathsf{\dfrac{1+sqrt{3}}{\sqrt{3}}=\dfrac{sqrt{3}+3}{3}}

o)

\mathsf{\dfrac{1-\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{2}-\sqrt{10}}{2}}

2)

a)

\mathsf{\dfrac{1}{\sqrt[5]{{2}^{4}}}=\dfrac{\sqrt[5]{2}}{2}}

b)

\mathsf{\dfrac{3}{\sqrt[3]{3}}}=\dfrac{3\sqrt[3]{{3}^{2}}}{3} =  \sqrt[3]{9}

c)

\mathsf{\dfrac{2a}{\sqrt[8]{{a}^{7}}}=\dfrac{2a\sqrt[8]{a}}{a}=2\sqrt[8]{a}}

d)

\mathsf{ \dfrac{2}{3 \sqrt[4]{ {2}^{3}}}=\dfrac{2\sqrt[4]{2} }{3.2}=\dfrac{\sqrt[4]{2} }{3}}

e)

\mathsf{\dfrac{ab}{ \sqrt[3]{ {a}^{2} {b}^{2}}}=\dfrac{ab \sqrt[3]{ {( {a}^{2} {b}^{2}  )}^{2} } }{ {a}^{2} {b}^{2}  } = \dfrac{ \sqrt[3]{ {{a}^{4} {b}^{4}} } }{ab}   }

f)

 \large\mathsf{ \dfrac{m}{ \sqrt[6]{ {n}^{5} } }  =  \dfrac{m \sqrt[6]{n} }{n} }


alic157: bgdddddd
CyberKirito: De nada
CyberKirito: Obg ^^.
luizfilipetimm: Me Mudem no meu
luizfilipetimm: Ajudem*
Perguntas interessantes