Cho x,y,z Thỏa mãn x+y+z=7 x^2+y^2+z^2=23 xyz=3 Tính H= $\frac{1}{xy+z-6}+$ $\frac{1}{yz+x-6}+$ $\frac{1}{xz+y-6}$

Question

Cho x,y,z
Thỏa mãn x+y+z=7
x^2+y^2+z^2=23
xyz=3
Tính H= $\frac{1}{xy+z-6}+$ $\frac{1}{yz+x-6}+$ $\frac{1}{xz+y-6}$

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King 5 years 2021-05-17T04:39:34+00:00 1 Answers 96 views 0

Answers ( )

  1. kiet
    0
    2021-05-17T04:41:21+00:00
    Reply

    Cho `x,y,z`

    Thỏa mãn `x+y+z=7`

                    `x^2+y^2+z^2=23`

                    `xyz=3`

    Tính $H=\dfrac{1}{xy+z-6}+\dfrac{1}{yz+x-6}+\dfrac{1}{xz+y-6}$

    Giải

    Ta có:` x+y+z=7`

    `⇔z=-x-y+7`

    `⇔xy+z-6=xy-x-y+7-6`

    `⇔xy+z-6=xy-x-y+1`

    `⇔xy+z-6=x(y-1)-(y-1)`

    `⇔xy+z-6=(y-1)(x-1)`

    Tương tự ta có:

    `yz+x-6=(y-1)(z-1)`

    `xz+y-6=(x-1)(z-1)`

    Ta lại có: 

    `(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)`

    `⇔7^2=23+2(xy+xz+yz)`

    $⇔xy+xz+yz=\dfrac{49-23}{2}=13$ 

    $H=\dfrac{1}{xy+z-6}+\dfrac{1}{yz+x-6}+\dfrac{1}{xz+y-6}$

    $⇔H=\dfrac{1}{(y-1)(x-1)}+\dfrac{1}{(y-1)(z-1)}+\dfrac{1}{(x-1)(z-1)}$

    $⇔H=\dfrac{z-1+x-1+y-1}{(y-1)(x-1)(z-1)}$ 

    $⇔H=\dfrac{z+x+y-3}{(xy-y-x+1)(z-1)}$ 

    $⇔H=\dfrac{7-3}{xyz-xy+y-yz-xz+x+z-1}$

    $⇔H=\dfrac{4}{(x+y+z)-(xy+xz+yz)+xyz-1}$

    $⇔H=\dfrac{4}{7-(xy+xz+yz)+3-1}$

    $⇔H=\dfrac{4}{9-(xy+xz+yz)}$

    $⇔H=\dfrac{4}{9-13}$

    $⇔H=\dfrac{4}{-4}=-1$

    Vậy `H=-1`

     

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