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प्रश्न
If `|("x"^"k", "x"^("k" + 2), "x"^("k" + 3)),("y"^"k", "y"^("k" + 2), "y"^("k" + 3)),("z"^"k", "z"^("k" + 2), "z"^("k" + 3))|` = (x - y) (y - z) (z - x)`(1/"x"+ 1/"y" + 1/"z") ` then
पर्याय
k = –3
k = –1
k = 1
k = 3
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उत्तर
k = – 1
Explanation:
L.H.S →
`|(x^k, x^(k + 2), x^(k + 3)),(y^k, y^(k + 2), y^(k + 3)),(z^k, z^(k + 2), z^(k + 3))| = x^k y^k z^k |(1, x^2, x^3),(1, y^2, y^3),(1, z^2, z^3)|`
= (xyz)k(x – y) (y – z) (z – x) (xy + yz + zx)
R.H.S. →
= `(x - y) (y - z) (z - x) (1/x+ 1/y + 1/z)`
= `(x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`
L.H.S. = R.H.S.
∴ `(xyz)^k (x - y) (y - z) (z - x) (xy + yz + zx) = (x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`
∴ `(xyz)^k cancel((x - y)) cancel((y - z)) cancel((z - x)) (xy + yz + zx) = cancel((x - y)) cancel((y - z)) cancel((z - x)) ((xy + yz + zx)/(xyz))`
∴ `(xyz)^k cancel((xy + yz + zx)) = cancel((xy + yz + zx))/(xyz)`
∴ (xyz)k = `1/"xyz"`
∴ (xyz)k = xyz- 1
∴ k = – 1
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