Cho $\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ =0 ( x, y, z $\neq$ 0). Tính $\frac{yz}{x^{2}}$ + $\frac{xz}{y^{2}}$ + $\frac{xy}{z^{2}}$

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Môn Toán RI SƠ 4 years 2020-10-14T18:27:03+00:00 2020-10-14T18:27:03+00:00 1 Answers 105 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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Euphemia · Answer 1 · 2020-10-14T18:28:11+00:00

Đáp án:

$\sum = 3$

Giải thích các bước giải:

Ta có:

$\dfrac{1}{x} +\dfrac{1}{y} + \dfrac 1z = 0$

$\to \dfrac{xy + yz + zx}{xyz} = 0$

$\to xy + yz + zx = 0$

$\to \left(\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2}\right)(xy + yz + zx) = 0$

$\to \dfrac{xy + yz + zx}{x^2} + \dfrac{xy + yz + zx}{y^2} + \dfrac{xy + yz + zx}{z^2} = 0$

$\to \dfrac yx + \dfrac{yz}{x^2} + \dfrac zx + \dfrac xy + \dfrac zy + \dfrac{zx}{y^2} + \dfrac{xy}{z^2} + \dfrac yz +\dfrac xz = 0$ $(*)$

Mặt khác:

$\dfrac{1}{x} +\dfrac{1}{y} + \dfrac 1z = 0$

$\to \begin{cases}1 + \dfrac xy + \dfrac xz = 0\\\dfrac yx + 1 + \dfrac yz = 0\\\dfrac zx + \dfrac zy + 1 = 0\end{cases}$

$\to \dfrac xy + \dfrac xz +\dfrac yx + \dfrac yz +\dfrac zx + \dfrac zy = -3$

Thay vào $(*)$ ta được:

$\dfrac{yz}{x^2} + \dfrac{zx}{y^2} + \dfrac{xy}{z^2} – 3 = 0$

$\to \dfrac{yz}{x^2} + \dfrac{zx}{y^2} + \dfrac{xy}{z^2} = 3$