Question 133

If x=y=z, then $$\ \dfrac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}}\ $$is:

Name: If x=y=z, then \ \dfrac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}}\ is:
Uploaded: March 30, 2021, 11:22 a.m.
Duration: 1 min 1 s
Description: If x=y=z, then \ \dfrac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}}\ is:

Solution

Given : $$x=y=z$$

Let $$x=y=z=k$$

To find : $$\ \dfrac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}}\ $$

= $$\dfrac{(k+k+k)^2}{k^2+k^2+k^2}$$

= $$\dfrac{(3k)^2}{3k^2}$$

= $$\dfrac{9k^2}{3k^2}=3$$

=> Ans - (B)

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SSC CHSL 15 November 2015 Evening Shift

If x=y=z, then $$\ \dfrac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}}\ $$is:

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