Question

[Nguyễn Thị Diệp]

Bài 1: Tìm GTNN

         N= 2x^2+4y^2-2x-4y+15

Bài 2:Tìm X

         a) (2x+3)^2-25(1-x)^2=0 

Answer 1

[Lấp La Lấp Lánh]

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:

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

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

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

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

Answer 2

[Nguyễn Thị Diệp]

Bạn vào giúp mk thêm câu nữa nhé.