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KR
7 tháng 5 2018
Áp dụng Bunyakovsky, ta có :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)
=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)
=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)
Mấy cái kia tương tự
M
1
HM
16 tháng 7 2023
c) C = x2 - 2xy+ y2 + y^2 - 2y +1 +2
= (x-y)^2 + (y - 1)^2 + 2
Ta có (x-y)^2;(y-1)^2 lớn hơn hoặc bằng 2
Dấu "=" xảy ra => (y-1)^2 = 0 => y-1=0 => y = 1
(x-y)^2 = 0 => x - y = 0 => x - 1= 0 => x = 1
d) D = \(4x^2-4x+1+x^2+2xy+y^2-2024\)
= \(\left(2x-1\right)^2+\left(x+y\right)^2-2024\)
Ta có (2x-1)^2;(x+y)^2 ≥ 0 => D ≥ -2024
Dấu = xảy ra => (2x-1)^2 = 0 => 2x-1 = 0 => x = 1/2
=> (x+y)^2 =0 => x+y=0 => 1/2+y =0 => y = -1/2
NK
1
DD
Đoàn Đức Hà
Giáo viên
6 tháng 11 2021
a) \(A=6x-x^2-11=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)
Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\).
b) \(B=x^2-5x-2=x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2-\frac{33}{4}=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge-\frac{33}{44}\)
Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).
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NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2021
Câu này em đã hỏi rồi
1.Tìm GTNN của Bthức : B= 4x2- 6x+1 : (x-2)2 với x ≠ 22. Tìm GTLN của Bthức: C= x2 + 4x - 14 : x2 -2x +1 với x≠ 1gi... - Hoc24
18 tháng 10 2021
a) \(A=4x^2-4x+9=\left(4x^2-4x+1\right)+8\)
\(=\left(2x-1\right)^2+8\ge8\)
\(minA=8\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=\left(x^2-4x+4\right)+\left(y^2+5y+\dfrac{25}{4}\right)-\dfrac{13}{4}\)
\(=\left(x-2\right)^2+\left(y+\dfrac{5}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\)
\(minC=-\dfrac{13}{4}\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=-\dfrac{5}{2}\end{matrix}\right.\)
\(=\left(x^2-2xy+y^2\right)-12\left(x-y\right)+36+5y^2+9\)
\(=\left(x-y\right)^2-12\left(x-y\right)+36+3y^2+9\)
\(=\left(x-y-6\right)^2+5y^2+9\ge9\)
Vậy GTNN của biểu thức là 9, xảy ra khi \(x=6;y=0\)