Question

36)Determine o valor de X nas igualdades

a)
$\sqrt{34 \: . \sqrt{8} } = 4 \sqrt{ \times }$
b)
$\sqrt{ \times .} \sqrt{40} = 8 \sqrt{10}$
c)
$\sqrt{ {4}^{ \times } } \div \sqrt{2} = 4 \sqrt{2}$
d)
$\sqrt{550} \div \sqrt{11} = \times \sqrt{12}$
​

Answer 1

Resposta:

a) $x=\frac{17\sqrt{2}}{4}$  ;  b) $x=16$  ;  c) $x=3$  ;  d) $x=\frac{5\sqrt{6}}{6}$

Explicação passo-a-passo:

a)  $\sqrt{34.\sqrt{8}}=4\sqrt{x}$

    $\sqrt{34.\sqrt{2^{3}}}=4\sqrt{x}$

    $\sqrt{34.2\sqrt{2}}=4\sqrt{x}$

    $\sqrt{68\sqrt{2}}=4\sqrt{x}$

    $\sqrt{2^{2}.17\sqrt{2}}=4\sqrt{x}$

    $2\sqrt{17\sqrt{2}}=4\sqrt{x}$

    $\frac{2\sqrt{17\sqrt{2}}}{4}=\sqrt{x}$

    $\frac{\sqrt{17\sqrt{2}}}{2}=\sqrt{x}$

    Eleve ambos os termos ao quadrado

    $(\frac{\sqrt{17\sqrt{2}}}{2})^{2}=(\sqrt{x})^{2}$

    $x=\frac{17\sqrt{2}}{4}$

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b)  $\sqrt{x}.\sqrt{40}=8\sqrt{10}$

    $\sqrt{x}=\frac{8\sqrt{10}}{\sqrt{40}}$

   

    Eleve ambos os termos ao quadrado

    $(\sqrt{x})^{2}=(\frac{8\sqrt{10}}{\sqrt{40}})^{2}$

    $x=\frac{64.10}{40}$

    $x=\frac{64}{4}$

    $x=16$

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c)  $\sqrt{4^{x}}:\sqrt{2}=4\sqrt{2}$

    $\frac{\sqrt{4^{x}}}{\sqrt{2}}=4\sqrt{2}$

    $\sqrt{4^{x}}=4\sqrt{2}.\sqrt{2}$

    $\sqrt{4^{x}}=8$

    Eleve ambos os termos ao quadrado

    $(\sqrt{4^{x}})^{2}=8^{2}$

    $4^{x}=64$

    $4^{x}=4^{3}$

    $x=3$

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d)  $\sqrt{550}:\sqrt{11}=x\sqrt{12}$

    $\frac{\sqrt{550}}{\sqrt{11}}=x\sqrt{12}$

    $x=\frac{\frac{\sqrt{550}}{\sqrt{11}}}{\sqrt{12}}$

    $x=\frac{\sqrt{550}}{\sqrt{11}}.\frac{1}{\sqrt{12}}$

    $x=\sqrt{\frac{550}{11}}.\frac{1}{\sqrt{12}}$

    $x=\sqrt{50}.\frac{1}{\sqrt{12}}$

    $x=\sqrt{5^{2}.2}.\frac{1}{\sqrt{2^{2}.3}}$

    $x=5\sqrt{2}.\frac{1}{2\sqrt{3}}$

    $x=\frac{5\sqrt{2}}{2\sqrt{3}}$

    $x=\frac{5\sqrt{2}}{2\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}$

    $x=\frac{5\sqrt{6}}{2.3}$

    $x=\frac{5\sqrt{6}}{6}$