(9x^2-6x+1)+y^2+4

=(3x-1)^2+y^2+4

ta có (3x-1)^2>= 0

=>(3x-1)^2+y^2>=0

=>(3x-1)^2+y^2+4>=4

GTNN biểu thức là 4 và xảy ra khi 3x-1=0=>x=1/3, y=0

`2/[6x-5-9x^2]`

`=-2/[9x^2-6x+5]`

`=-2/[(3x-1)^2+4]`

Vì `(3x-1)^2 >= 0 AA x`

`<=>(3x-1)^2+4 >= 4 AA x`

`<=>1/[(3x-1)^2+4] <= 1/4`

`<=>-2/[(3x-1)^2+4] >= -1/2 AA x`

   `=>Mi n=-1/2`

Dấu "`=`" xảy ra `<=>3x-1=0<=>x=1/3`

Ta có: \(D=9x^2+12x-1=9x^2+12x+4-3=\left(3x+2\right)^2-3\)

Mà: \(D=\left(3x+2\right)^2-3\le-3\forall x\) 

Dấu "=" xảy ra 

\(\Leftrightarrow\left(3x+2\right)^2=0\Leftrightarrow3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy \(D_{min}=-3\Leftrightarrow x=-\dfrac{2}{3}\)

\(A=2x^2+8x-24\)

\(=2\left(x^2+4x-12\right)\)

\(=2\left[x^2+4x-4-8\right]\)

\(=2\left[\left(x-2\right)^2-8\right]\)

\(\left(x-2\right)^2\ge0\)

\(\Rightarrow\left(x-2\right)^2-8\ge-8\)

\(\Rightarrow2\left[\left(x-2\right)^2-8\right]\ge-16\)

Do đó GTNN của A là -16 khi \(x-2=0\Rightarrow x=2\)

\(B=x^2-8x+5=x^2-8x+16-9\)

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

\(=\left(x-4\right)^2-9\)

\(\left(x-4\right)^2\ge0\)

\(\Rightarrow\left(x-4\right)^2-9\ge-9\)

Do đó GTNN của B là -9 khi \(x-4=0\Rightarrow x=4\)

A=4x^2+8x+4+1

=(2x+2)^2+1>=1

Dấu = xảy ra khi x=-1

\(a,=x^2-8x+16+1=\left(x-4\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=4\)

\(b,=\left(4x^2-12x+9\right)+4=\left(2x-3\right)^2+4\ge4\)

Dấu \("="\Leftrightarrow x=\dfrac{3}{2}\)

\(c,=\left(9x^2-2\cdot3\cdot\dfrac{1}{3}x+\dfrac{1}{9}\right)+\dfrac{26}{9}=\left(3x-\dfrac{1}{3}\right)^2+\dfrac{26}{9}\ge\dfrac{26}{9}\)

Dấu \("="\Leftrightarrow3x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{9}\)

a) = 3(x²-2x+1) +1-3

GTNN = -2

B) tt

\(C=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\left(x< 0\right)\)

\(D=x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)

\(=x-4+\sqrt{\left(x-4\right)^2}\)

\(=x-4+\left|x-4\right|\)

\(=x-4+x-4\)

\(=2x-8\)

\(C=\sqrt{9x^2}-2x=-3x-2x=-5x\)

\(D=x-4+\sqrt{x^2-8x+16}=x-4+x-4=2x-8\)

1) \(A=x^2+8x+15=\left(x^2+8x+16\right)-1=\left(x+4\right)^2-1\ge-1\)

\(minA=-1\Leftrightarrow x=-4\)

2) \(B=7x-x^2-5=-\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{29}{4}=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{29}{4}\le\dfrac{29}{4}\)

\(maxB=\dfrac{29}{4}\Leftrightarrow x=\dfrac{7}{2}\)

Ta có: \(A=x^2+8x+15\)

\(=x^2+8x+16-1\)

\(=\left(x+4\right)^2-1\ge-1\forall x\)

Dấu '=' xảy ra khi x=-4

Lớp 8 nhé, mình chọn nhầm