Bài 2:
Ta có: \(E=x^2+2y^2-2xy-4y+7\)
\(=x^2-2xy+y^2+y^2-4y+4+3\)
\(=\left(x-y\right)^2+\left(y-2\right)^2+3\ge3\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}x-y=0\\ y-2=0\end{cases}\Rightarrow x=y=2\)
Bài 1:
a: \(\left(2x-1\right)^2\ge0\forall x\)
=>\(-\left(2x-1\right)^2\le0\forall x\)
=>\(-\left(2x-1\right)^2+2025\le2025\forall x\)
Dấu '=' xảy ra khi 2x-1=0
=>2x=1
=>\(x=\frac12\)
b: \(B=4x-x^2+17\)
\(=-\left(x^2-4x-17\right)\)
\(=-\left(x^2-4x+4-21\right)\)
\(=-\left(x-2\right)^2+21\le21\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
c: \(C=\frac{1}{4x^2+4x+9}\)
\(=\frac{1}{4x^2+4x+1+8}\)
\(=\frac{1}{\left(2x+1\right)^2+8}\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
=>\(\left(2x+1\right)^2+8\ge8\forall x\)
=>\(C=\frac{1}{\left(2x+1\right)^2+8}\le\frac18\forall x\)
Dấu '=' xảy ra khi 2x+1=0
=>2x=-1
=>\(x=-\frac12\)
d: \(D=\frac{5x^2+21}{x^2+3}\)
\(=\frac{5x^2+15+6}{x^2+3}=5+\frac{6}{x^2+3}\)
Ta có: \(x^2+3\ge3\forall x\)
=>\(\frac{6}{x^2+3}\le\frac63=2\forall x\)
=>\(D=\frac{6}{x^2+3}+5\le2+5=7\forall x\)
Dấu '=' xảy ra khi x=0
Bài 1: a) A = 2025 - (2x - 1)^2. Vì (2x - 1)^2 ≥ 0 voi moi x, gia tri nho nhat cua no la 0 khi 2x - 1 = 0 (x = 0,5). Do do A dat gia tri lon nhat la 2025 tai x = 0,5. b) B = -x^2 + 4x + 17. Day la ham bac hai co he so a = -1 < 0 nen co cuc dai tai x = -b/(2a) = -4/(2* -1) = 2. Thay x = 2 vao B, ta co B_max = -4 + 8 + 17 = 21. c) C = 1/(4x^2 + 4x + 9). Mau so 4x^2 + 4x + 9 luon duong va dat gia tri nho nhat khi x = -b/(2a) = -4/8 = -0,5. Khi do 4x^2 + 4x + 9 = 8 nen C_max = 1/8. d) D = (5x^2 + 21)/(x^2 + 3) = 5 + 6/(x^2 + 3). Vi x^2 + 3 ≥ 3, phan thuc 6/(x^2 + 3) dat lon nhat khi x = 0. Do do D_max = 5 + 6/3 = 7 tai x = 0. Bài 2: E = x^2 + 2y^2 - 2xy + 7 = (x - y)^2 + y^2 + 7. Cac so binh phuong deu khong am nen gia tri nho nhat cua E la 7, dat duoc khi x = y = 0.
