જો y^2+z^2=3yz,z^2+x^2=8zx,અને x^2+y^2=4xy હો | vedclass

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(C) આપેલ સમીકરણો:$y^2+z^2=3yz \Rightarrow \frac{y}{z}+\frac{z}{y}=3$  $(i)$$z^2+x^2=8zx \Rightarrow \frac{z}{x}+\frac{x}{z}=8$  $(ii)$$x^2+y^2=4xy \Ri

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) આપેલ સમીકરણો:$y^2+z^2=3yz \Rightarrow \frac{y}{z}+\frac{z}{y}=3$  $(i)$$z^2+x^2=8zx \Rightarrow \frac{z}{x}+\frac{x}{z}=8$  $(ii)$$x^2+y^2=4xy \Rightarrow \frac{x}{y}+\frac{y}{x}=4$  $(iii)$$(i)$ અને $(iii)$ નો ગુણાકાર કરતા:$\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right) = \frac{x}{z} + \frac{y^2}{xz} + \frac{xz}{y^2} + \frac{z}{x} = 12$$\left(\frac{x}{z}+\frac{z}{x}\right) + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$$(ii)$ પરથી કિંમત મૂકતા:$8 + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$$\frac{y^2}{xz}+\frac{xz}{y^2} = 12 - 8 = 4$

જો $y^2+z^2=3yz$,$z^2+x^2=8zx$,અને $x^2+y^2=4xy$ હોય,તો $\frac{y^2}{xz}+\frac{xz}{y^2}$ ની કિંમત શોધો.

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