Toán Lớp 8: Tìm Min hoặc Max
H=(2x^2 – 2xy + 9y^2)/(x^2+2xy+5y^2)
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Toán Lớp 8: Tìm Min hoặc Max
H=(2x^2 – 2xy + 9y^2)/(x^2+2xy+5y^2)
Comments ( 1 )
17/4-H
=17/4 – (2x^2-2xy + 9y^2)/(x^2+2xy+5y^2)
= (17x^2 + 34xy + 85y^2 – 8x^2 + 8xy- 36y^2)/(4 (x^2+2xy+5y^2))
= (9x^2 + 42xy +49y^2)/(4(x^2+2xy + y^2+4y^2))
= (3x+7y)^2/(4(x+y)^2+4y^2)\ge 0
->17/4-H \ge 0->H\le 17/4
Dấu “=” xảy ra khi : 3x+7y=0<=>x=(-7y)/3
Vậy max H=17/4<=>x=(-7y)/3
H-1
=(2x^2-2xy + 9y^2)/(x^2+2xy+5y^2) – (x^2+2xy+5y^2)/(x^2+2xy+5y^2)
=(x^2 – 4xy + 4y^2)/(x^2+2xy+y^2+4y^2)
=(x-2y)^2/((x+y)^2+4y^2)\ge 0
->H-1\ge 0->H\ge 1
Dấu “=” xảy ra khi : x-2y=0<=>x=2y
Vậy min H=1<=>x=2y