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

GTNN = 5

2) tuong tu 

\(D=\left(x^4-2x^3+x^2\right)+\left(2x^2-2x+1\right)\)

\(D=\left(x^2-x\right)^2+2\left(x^2-x\right)+1=\left(x^2-x+1\right)^2\)

\(D=\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]^2\)

\(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow D\ge\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)

đẳng thúc khi x=1/2

{logic 10x-->10x^2}

\(E=x^4-6x^3+10x^2-6x+9\)

\(E=\left(x^4-3x+9x^2\right)+\left(x^2-6x+9\right)\)

\(E=\left(x^2-3x\right)^2+\left(x-3\right)^2=\left[x^2\left(x-3\right)^2\right]+\left(x-3\right)^2\)

\(E=\left(x-3\right)^2\left(x^2+1\right)\)

\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(x^2+1\right)\ge1\end{matrix}\right.\) \(\Rightarrow E\ge0\) đẳng thức khi x=3

a) \(=\left(9x^2+2.3.\frac{5}{3}x+\frac{25}{9}\right)-\frac{34}{9}=\left(3x+\frac{5}{3}\right)^2-\frac{34}{9}\ge-\frac{34}{9}\Rightarrow Min=-\frac{34}{9}\Leftrightarrow x=-\frac{5}{9}\)

b) \(=2\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\Rightarrow Min=-\frac{9}{2}\Leftrightarrow x=\frac{3}{2}\)

A=(x²+2.x².4+4²)+4=(x+4)²+4 =>gtnn của A là 4 tại x=-4

câu dưới tương tự nhưng đặt nhân tử chung là 2 ra ngoài nha

A=x²+8x+20

=x²+8x+16+4

=(x+4)²+4\(\ge\)0+4=4

Dấu = khi x+4=0 <=>x=-4

Vậy Amin=4 khi x=-4

B=2x²+10x+20

\(=2\left(x^2+\frac{10x}{2}+10\right)\)

\(=2\left(x^2+\frac{5x}{2}+\frac{5x}{2}+\frac{25}{4}\right)+\frac{15}{2}\)

\(=2\left(x+\frac{5}{2}\right)^2+\frac{15}{2}\ge0+\frac{15}{2}=\frac{15}{2}\)

Dấu = khi x+5/2=0 <=>x=-5/2

Vậy Bmin=15/2 khi x=-5/2

\(P=2x^2-10x+13=2\left(x^2-5x+\frac{25}{4}\right)+\frac{1}{2}\)

                                            \(=2\left(x-\frac{5}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Dấu "="  xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy \(P_{min}=\frac{1}{2}\Leftrightarrow x=\frac{5}{2}\)

\(P=2x^2-10x+13\)

\(P=2\left(x^2-5x+\frac{13}{2}\right)\)

\(P=2\left[\left(x^2-2.x.\frac{5}{2}+\frac{25}{4}\right)-\frac{1}{4}\right]\)

\(P=2\left[\left(x-\frac{5}{2}\right)^2-\frac{1}{4}\right]\)

\(P=2\left(x-\frac{5}{2}\right)^2-\frac{1}{2}\ge\frac{1}{2}\)

\(\Rightarrow Pmin=\frac{-1}{2}\Leftrightarrow x=\frac{5}{2}\)

\(A=2x^2+4y^2+4xy+10x+12y+18\)

\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)

\(A=\left(x+2y^2\right)+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)

\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)

Do \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)

\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)

" = " \(\Leftrightarrow\hept{\begin{cases}x+2y+3=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=-2\end{cases}}}\)

\(\Rightarrow A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)

Chúc bạn học tốt !!!