Cho 1/x+1/y-1/z=0. Tính S=xy/z^2-yz/x^2-zx/y^2

0 bình luận về “Cho 1/x+1/y-1/z=0. Tính S=xy/z^2-yz/x^2-zx/y^2”

1. Đáp án:

   $S=3$

   Giải thích các bước giải:

    ĐKXĐ: $x,y,z\ne 0$

   Ta có:

   $\dfrac{1}{x} + \dfrac{1}{y} – \dfrac{1}{z} = 0 \Rightarrow \left\{ \begin{array}{l}
   \dfrac{1}{z} = \dfrac{1}{x} + \dfrac{1}{y}\\
   \dfrac{1}{x} = \dfrac{1}{z} – \dfrac{1}{y}\\
   \dfrac{1}{y} = \dfrac{1}{z} – \dfrac{1}{x}
   \end{array} \right.$

   Khi đó:

   $\begin{array}{l}
   S = \dfrac{{xy}}{{{z^2}}} – \dfrac{{yz}}{{{x^2}}} – \dfrac{{zx}}{{{y^2}}}\\
    = xy.{\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right)^2} – yz.{\left( {\dfrac{1}{z} – \dfrac{1}{y}} \right)^2} – zx.{\left( {\dfrac{1}{z} – \dfrac{1}{x}} \right)^2}\\
    = xy\left( {\dfrac{1}{{{x^2}}} + \dfrac{2}{{xy}} + \dfrac{1}{{{y^2}}}} \right) – yz\left( {\dfrac{1}{{{z^2}}} – \dfrac{2}{{yz}} + \dfrac{1}{{{y^2}}}} \right) – zx\left( {\dfrac{1}{{{z^2}}} – \dfrac{2}{{xz}} + \dfrac{1}{{{x^2}}}} \right)\\
    = \dfrac{y}{x} + 2 + \dfrac{x}{y} – \dfrac{y}{z} + 2 – \dfrac{z}{y} – \dfrac{x}{z} + 2 – \dfrac{z}{x}\\
    = 6 + \left( {\dfrac{y}{x} – \dfrac{y}{z}} \right) + \left( {\dfrac{x}{y} – \dfrac{x}{z}} \right) – \left( {\dfrac{z}{y} + \dfrac{z}{x}} \right)\\
    = 6 + y\left( {\dfrac{1}{x} – \dfrac{1}{z}} \right) + x\left( {\dfrac{1}{y} – \dfrac{1}{z}} \right) – z\left( {\dfrac{1}{y} + \dfrac{1}{x}} \right)\\
    = 6 + y.\left( {\dfrac{{ – 1}}{y}} \right) + x.\left( {\dfrac{{ – 1}}{x}} \right) – z.\dfrac{1}{z}\\
    = 6 – 1 – 1 – 1\\
    = 3
   \end{array}$

   Vậy $S=3$

   Bình luận