Question 133
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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