\(A=3x^2+6x+15=3\left(x^2+2x+1\right)+12\)

\(=3\left(x+1\right)^2+12\ge12\)

\(minA=12\Leftrightarrow x=-1\)

\(a,=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)

A=(2x)²+2.2x.\(\frac{7}{4}\)+\(\frac{49}{16}-\frac{49}{16}+13\)=(2x+7/2)²+159/16

ta co (2x+7/2)²> hoac bang 0

--> (2x+7/2)²+159/16> hoac bang 159/16

vay A dat gia tri nn la 159/16 khi x=-7/4

B= x²-2x.4+16-16+5=(x-4)²-11

ta co (x-4)² > hoac bang 0

==> (x-4)²-11> hoac bang -11

vay B dat gtnn la -11 khi x=4

A = 2x² + 6x - 1

A = 2( x² + 3x - 1 / 2 )

A = 2[ x² + 2 . 3 / 2 . x + ( 3 / 2 )² - ( 3 / 2 )² - 1 / 2 ]

A = 2[ ( x + 3 / 2 )² - 11 / 4 ]

A = ( x + 3 / 2 )² - 11 / 2 \(\ge\)- 11 / 2

Dấu " = " xảy ra\(\Leftrightarrow\)x + 3 / 2 = 0

                            \(\Rightarrow\)x              = 3 / 2

Min A = - 11 / 2 \(\Leftrightarrow\)x = 3 / 2

Đặt A 9x² + 6x - 1 = 9x² + 6x + 1 - 2 = (3x + 1)² - 2 \(\ge\)-2

=> Min A = -2

Dấu "=" xảy ra <=> 3x + 1 = 0 

<=> x = -1/3

Vậy Min A = -2 <=> x = -1/3 

=-1/3 nha

Trả lời:

1, \(9x^2+6x-1=9x^2+6x+1-2=\left(3x+1\right)^2-2\ge-2\forall x\)

Dấu "=" xảy ra khi 3x + 1 = 0 <=> x = - 1/3

Vậy GTNN của bt = - 2 khi x = - 1/3

1: Ta có: \(x^2-2x-5\)

\(=x^2-2x+1-6\)

\(=\left(x-1\right)^2-6\ge-6\forall x\)

Dấu '=' xảy ra khi x=1

2: ta có: \(3x^2+5x-2\)

\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{2}{3}\right)\)

\(=3\left(x^2+2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)

\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{49}{12}\ge-\dfrac{49}{12}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{6}\)