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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Cho x,y,z >0 thỏa mãn `1/x`+`1/z` =`2021/y`. Tìm min : `P` = `(x+y)/(2021x-y)` + `(y+z)/(2021z-y)`

Toán Lớp 8: Cho x,y,z >0 thỏa mãn 1/x+1/z =2021/y. Tìm min :
P = (x+y)/(2021x-y) + (y+z)/(2021z-y)

Comments ( 1 )

  1. $\\$
    1/x+1/z=2021/y
    <=> 1/x + 1/z = 2021 . 1/y
    <=> (1/x + 1/z) . y = 2021
    <=> y/x + y/z = 2021
    Để đơn giản ta đặt y/x= k, y/z=h(k,h>0)
    <=> k+h=y/x + y/z = y (1/x+1/z) = y . 2021/y
    <=>k+h=2021
    P=(x+y)/(2021x-y) + (y+z)/(2021z-y)
    = ((x+y) . 1/x)/((2021x-y). 1/x) + ((y+z) . 1/z)/((2021z -y) . 1/z)
    = (1 + y/x)/(2021 – y/x) + (y/z + 1)/(2021 – y/z)
    = (1 + k)/(2021-k) + (h + 1)/(2021-h)
    = (k – 2021 + 2022)/(2021-k) + (h-2021+2022)/(2021-h)
    = (-(2021-k))/(2021-k) + (-(2021-h))/(2021-h) + 2022/(2021-k) + 2022/(2021-h)
    = -1 -1 + 2022 (1/(2021-k) + 1/(2021-h))
    =-2+2022 (1/(2021-k) + 1/(2021-h))
    Áp dụng BĐT 1/x+1/y\ge 4/(x+y) (x,y>0) ta được :
    1/(2021-k) + 1/(2021-h) \ge 4/(2021-k + 2021-h) = 4/2021
    <=> 2022 (1/(2021-k) + 1/(2021-h)) \ge 2022 . 4/2021 = 8088/2021
    <=> P\ge -2 + 8088/20221=4046/2021
    Dấu “=” xảy ra khi :
    h=k <=> y/x=y/z
    <=>2 y/z = 2021
    <=> y/z=2021/2
    <=> x/y = y/z =2021/2
    Vậy min P=4046/2021<=>x/y=y/z=2021/2

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222-9+11+12:2*14+14 = ? ( )