Toán Lớp 8: Tìm Min hoặc Max
Q=(5y^2-3xy)/(x^2-3xy+4y^2)
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: Tìm Min hoặc Max
Q=(5y^2-3xy)/(x^2-3xy+4y^2)
Comments ( 1 )
9/7 – Q
= 9/7 – (5y^2-3xy)/(x^2-3xy+4y^2)
= (9x^2 – 27xy + 36y^2 – 35y^2 + 21xy)/(7 (x^2-3xy + 9/4y^2 + 7/4y^2))
= (9x^2 – 6xy + y^2)/(7 ((x-3/2y)^2 + 7/4y^2))
= (3x-y)^2/(7((x-3/2y)^2 + 7/4y^2))\ge 0
-> 9/7-Q\ge 0 ->Q\le 9/7
Dấu “=” xảy ra khi :3x-y=0<=>3x=y
Vậy max Q=9/7<=>3x=y
1+Q
= 1+(5y^2-3xy)/(x^2-3xy + 4y^2)
= (x^2-3xy + 4y^2 +5y^2-3xy)/(x^2-3xy+4y^2)
= (x^2 – 6xy + 9y^2)/(x^2-3xy + 9/4 y^2 + 7/4y^2)
= (x-3y)^2/((x-3/2y)^2 + 7/4y^2)\ge 0
-> 1+Q\ge 0->Q\ge -1
Dấu “=” xảy ra khu : x-3y=0<=>x=3y
Vậy min Q=-1<=>x=3y