Math, asked by laliteshwarkumar, 1 year ago

5(1/x2+1/y2+1/z^2)=4(1/xy+1/yz+1/zx)

Answers

Answered by nobel
2
Algebra,

We have,

5 \times (\frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} } + \frac{1}{ {z}^{2} } ) = 4 \times ( \frac{1}{xy} + \frac{1}{yz} + \frac{1}{xz} )

1/x²+ 1/y²+ 1/z²= 4/5(1/xy + 1/yz + 1/xz)

(1/x + 1/y + 1/z)²-2(1/xy + 1/yz + 1/xz) = 4/5(1/xy + 1/yz + 1/xz)

(1/x + 1/y + 1/z)²=4/5(1/xy + 1/yz + 1/xz) + 2(1/xy + 1/yz + 1/xz)

(1/x + 1/y + 1/z)²=14/5(1/xy + 1/yz + 1/xz)

so the ansewr is,
 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \sqrt{\frac{14}{5} ( \frac{1}{xy} + \frac{1}{yz} + \frac{1}{xz})}

or, 1/x + 1/y + 1/z = 1/xyz√{14/5(xy + yz + xz)}

I think it's the correct, infact sure it's the answer.

That's it
Hope it helped (●´ϖ`●)
Attachments:

laliteshwarkumar: LALITESHWAR KUMAR ( PRABHAT KUMAR ):
5(1/x² + 1/y² + 1/z²) = 4(1/xy + 1/yz + 1/zx)

4/x² + 4/y² + 4/z² + 1/x² + 1/y² + 1/z² - 4/xy - 4/yz - 4/zx = 0

(4/x² + 1/y² - 4/xy) + (4/y² + 1/z² - 4/yz) + (4/z² + 1/x² - 4/zx) = 0

(2/x - 1/y)² + (2/y - 1/z)² + (2/z - 1/x)² = 0

So,,
2/x - 1/y = 0
2/y - 1/z = 0
2/z - 1/x = 0
----------Add these ------------
2(1/x + 1/y + 1/z) - 1/x + 1/y + 1/z = 0

1/x + 1/y + 1/z = 0 ...Answer
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