Bài 1:

\(N=2x^2+4y^2-2x-4y+15=2\left(x^2-x+\dfrac{1}{4}\right)+\left(4y^2-4y+1\right)+\dfrac{27}{2}=2\left(x-\dfrac{1}{2}\right)^2+\left(2y-1\right)^2+\dfrac{27}{2}\ge\dfrac{27}{2}\)

\(minN=\dfrac{27}{2}\Leftrightarrow x=y=\dfrac{1}{2}\)

Bài 2:

\(\Leftrightarrow4x^2+12x+9-25x^2+50x-25=0\)

\(\Leftrightarrow21x^2-62x+16=0\)

\(\Leftrightarrow\left(3x-8\right)\left(7x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{2}{7}\end{matrix}\right.\)

1) (x-1)² + (x- 4y)² + (y + 2)² +10 -1-4

GTNN = 5

2) tuong tu 

a/ \(x+4y=1\Rightarrow x=1-4y\)

\(A=x^2+4y^2=\left(1-4y\right)^2+4y^2=20y^2-8y+1\)

\(A=20\left(y^2-2.\frac{1}{5}y+\frac{1}{25}\right)+\frac{1}{5}=20\left(y-\frac{1}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}\)

\(\Rightarrow A_{min}=\frac{1}{5}\) khi \(\left\{{}\begin{matrix}y=\frac{1}{5}\\x=1-4y=\frac{1}{5}\end{matrix}\right.\)

b/

\(B=\frac{2x^2+5x+8}{x}=2x+\frac{8}{x}+5\ge2\sqrt{2x.\frac{8}{x}}+5=13\)

\(\Rightarrow B_{min}=13\) khi \(x=2\)

d)  D = x⁴ - 6x² + 10

D = (X²)² - 2. x². 3 + 3² + 1

D = (x² - 3)² + 1

(x² - 3)²  >= 0 với mọi x

(x² - 3)² + 1 >=1 với moi5 x

Vậy GTNN của D là 1

TÌM MIN NHÉ

trịnh phương anh mấy loại bạn

min là từ viết tắt của gtnn