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Question Number 61537 by aanur last updated on 04/Jun/19

Commented by aanur last updated on 04/Jun/19

$$sircouldyouhelpme$$

Commented by math1967 last updated on 04/Jun/19

$$xyz\times \sqrt{x^2\times y^2\times z^2}=4\times 9\times 16$$$$x^2\times y^2\times z^2=4\times 9\times 16\Rightarrow xyz=2\times 3\times 4$$

Answered by MJS last updated on 04/Jun/19

$$\left(1\right)\Rightarrow z=\frac{16}{x^{2}y}$$$$\left(2\right)\frac{4\sqrt{y}}{\sqrt{x}}=9$$$$\left(3\right)\frac{16}{\sqrt{x^{3}y}}=16$$$$$$$$\left(2\right)\Rightarrow y=\frac{81x}{16}\Rightarrow z=\frac{256}{81x^3}$$$$\left(3\right)\frac{64}{9x^2}=16\Rightarrow x=\frac{2}{3}\Rightarrow y=\frac{27}{8}\wedge z=\frac{32}{3}$$

Answered by Kunal12588 last updated on 04/Jun/19

$$i)x\sqrt{yz}=4$$$$ii)y\sqrt{xz}=9$$$$iii)z\sqrt{xy}=16$$$$i\times ii\times iii$$$$\Rightarrow xyz\sqrt{x^2{y}^2{z}^2}=2^6\times 3^2$$$$\Rightarrow {\left(xyz\right)}^2=2^6\times 3^2$$$$\Rightarrow xyz=2^3\times 3=24$$$${\left(i\right)}^2$$$$\Rightarrow x^{2}yz=4^2=16$$$$\Rightarrow \frac{x^{2}yz}{xyz}=\frac{16}{24}$$$$\Rightarrow x=\frac{2}{3}$$$${\left(ii\right)}^2$$$$\Rightarrow {xy}^{2}z=81$$$$\Rightarrow y=\frac{81}{24}=\frac{27}{8}$$$${\left(iii\right)}^2$$$$\Rightarrow {xyz}^2=256$$$$\Rightarrow z=\frac{256}{24}=\frac{32}{3}$$

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