Question

Determine o valor de X nas igualdades.A) Raiz de 34 . Raiz de 8 = 4 raiz de XB) Raiz de X . Raiz de 40 = 8 raiz de 10C) Raiz de 4 elevado a X : raiz de 2 = 4 raiz de 2D) raiz de 550 : raiz de 11 = X raiz de 2Me ajudem pfv !!!

Answer 1

A)
$\sqrt{34}* \sqrt{8} =4* \sqrt{x} \\ \sqrt{34*8}= \sqrt{x*4^2}\\ \sqrt{272}= \sqrt{16x}\\272=16x\\\frac{272}{16}=x\\x=17$

B)
$\sqrt{x} * \sqrt{40} =8* \sqrt{10} \\ \sqrt{x*40}= \sqrt{10*8^2} \\ \sqrt{40x} = \sqrt{640} \\40x=640\\x=\frac{640}{40}\\x=16$

C)
$\frac{\sqrt{4^x}}{\sqrt{2}}=4*\sqrt{2}\\\sqrt{\frac{(2^2)^x}{2}}=\sqrt{2*4^2}\\\sqrt{\frac{2^{2x}}{2}}=\sqrt{2*(2^2)^2}\\\sqrt{2^{2x}*2^{-1}}=\sqrt{2*2^4}\\\sqrt{2^{2x-1}}=\sqrt{2^5}\\2^{2x-1}=2^5\\2x-1=5\\2x=5+1\\2x=6\\x=\frac{6}{3}\\x=3$

D)
$\frac{\sqrt{550}}{\sqrt{11}}=x*\sqrt{2}\\\sqrt{\frac{550}{11}}=\sqrt{2*x^2}\\\sqrt{55}=\sqrt{2x^2}\\55=2x^2\\\frac{55}{2}=x^2$

$x' = \sqrt{\frac{55}{2}}\\\\x'' =- \sqrt{\frac{55}{2}}$