cho 1/2x+1/3y+1/z=0 tính S = 6xy/z^2 +3yz/4x^2+2zx/9y^2

Question

cho 1/2x+1/3y+1/z=0
tính S = 6xy/z^2 +3yz/4x^2+2zx/9y^2

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Ximena 30 phút 2021-10-16T14:53:44+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-16T14:54:50+00:00

    `S=(6xy)/(z^2)+(3xy)/(4x^2)+(2zx)/(9y^2)`

    `S=1/(2x)+1/(2y)+1/z=0<=>1/((2x)^2)+1/((3y)^2)+1/((z)^3)+1/(6xyz)`

    `S=6xyz[1/((2x)^2)+1/((3y)^2)+1/((z)^3)]`

    `S=6xyz.“3/(6xyz)`

    `S=3`

    Vậy `S={3}`

    0
    2021-10-16T14:55:35+00:00

    Đáp án: 3

     

    Giải thích các bước giải:

    Ta có $\frac{1}{2x}$ +$\frac{1}{3y}$ +$\frac{1}{z}=0$ $\Rightarrow$ $\frac{1}{(2x)^3}$+$\frac{1}{(3y)^3}$+$\frac{1}{(z)^3}$=$\frac{3}{6xyz}$ $\\ S=6xyz[\frac{1}{(2x)^3}+\frac{1}{(3y)^3}+\frac{1}{(z)^3}]$ $\\=6xyz. \frac{3}{6xyz}=3$ 

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