Question

x+y+z=0,3y²+x²+z²/2y²-xz=?​

Answer 1

Step-by-step explanation:

Givenx+y+z=0−−−(1)

\begin{gathered} LHS=\frac{x^{2}}{yz}+\frac{y^{2}}{zx}+\frac{z^{2}}{xy}\\=\frac{x^{3}+y^{3}+z^{3}}{xyz}\\=\frac{(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)+3xyz}{xyz}\\=\frac{0\times (x^{2}+y^{2}+z^{2}-xy-yz-zx)+3xyz}{xyz}\end{gathered}

LHS=

yz

x

2

+

zx

y

2

+

xy

z

2

=

xyz

x

3

+y

3

+z

3

=

xyz

(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)+3xyz

=

xyz

0×(x

2

+y

2

+z

2

−xy−yz−zx)+3xyz

/* From (1) */

=\frac{3xyz}{xyz}=

xyz

3xyz

\begin{gathered}= 3\\=RHS\end{gathered}

=3

=RHS

Therefore.,

\begin{gathered} If\: x+y+z=0 \: then \\\frac{x^{2}}{yz}+\frac{y^{2}}{zx}+\frac{z^{2}}{xy}=3\end{gathered}

Ifx+y+z=0then

yz

x

2

+

zx

y

2

+

xy

z

2

=3

•••♪