(x-1)(x-2)(x-3)(x-4)+15

=(x²-5x+4)(x²-5x+6)+15

Đặt t=x²-5x+4 ta có:

t(t+2)+15=t²+2t+15

=t²+2t+1+14=(t+1)²+14\(\ge\)14

Dấu = khi t=-1 => x²-5x+4=-1 =>x=\(\frac{5\pm\sqrt{5}}{2}\)

Vậy....

\(a,\) Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2+3\ge3\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy \(GTNN\) của đa thức là \(3\) khi \(x=1.\)

\(b,\) Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow-x^2\le0\forall x\)

\(\Rightarrow1-x^2\le1\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x^2=0\Leftrightarrow x=0\)

#\(Toru\)

\(A=4-x^2+3\)

\(=-x^2+7\le7\)

Khi x=0

\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

Đặt \(t=x^2+5x+4\) thì

\(=t\left(t+2\right)=t^2+2t+1-1\)

\(=\left(t+1\right)^2-1\ge-1\)

Khi x=0

Đặt  thì

Đặt:     \(A=\left(x-3\right)\left(x+3\right)+2\left(2x+1\right)^2\)

=>       \(A=x^2-9+2\left(4x^2+4x+1\right)\)

=>       \(A=x^2-9+8x^2+8x+2\)

=>       \(A=9x^2+8x-7\)

=>       \(A=\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\)

Có:      \(\left(3x+\frac{4}{3}\right)^2\ge0\forall x\Rightarrow\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\ge-\frac{79}{9}\)

=>      \(A\ge-\frac{79}{9}\)

DẤU "=" XẢY RA <=>     \(\left(3x+\frac{4}{3}\right)^2=0\)

<=>     \(x=-\frac{4}{9}\)

Vậy A min =     \(-\frac{79}{9}\)      <=>       \(x=-\frac{4}{9}\)

( x - 3 )( x + 3 ) + 2( 2x + 1 )²

= x² - 9 + 2( 4x² + 4x + 1 )

= x² - 9 + 8x² + 8x + 2

= 9x² + 8x - 7

= 9x² + 8x + 16/9 - 79/9

= ( 3x + 4/3 )² - 79/9

\(\left(3x+\frac{4}{3}\right)^2\ge0\forall x\Rightarrow\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\ge-\frac{79}{9}\)

Dấu " = " xảy ra <=> 3x + 4/3 = 0 => x = -4/9

=> GTNN của biểu thức = -79/9 <=> x =  -4/9

B=2(x^2+2.x.1/4 +1/16)^2 -57/8

  =2.(x+1/4)^2 -57/8

MinB=-57/8 khi x=-1/4

\(B=-14+2x^2+x=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{113}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{113}{8}\ge-\dfrac{113}{8}\)\(ĐTXR\Leftrightarrow x=-\dfrac{1}{4}\)

`B=x^2 +y^2 -2x+4y+2010`

`=x^2 -2x+1+y^2 +4y+4+2005`

`=(x-1)^2 + (y+2)^2 +2005 >= 2005`

Dấu "=" xảy ra `<=>{(x-1=0),(y+2=0):}<=>{(x=1),(y=-2):}`

Vậy `B_(min) = 2005 <=> {(x=1),(y=-2):}`

`B=x^2+y^2-2x+4y+2010`

`B=x^2-2x+y^2+4y+2010`

`B= x^2-2.x.1+1^2-1^2 +y^2+2y.2+2^2-2^2+2010`

`B= (x^2-2x+1)+(y^2+4y+4)-1-4+2010`

`B= (x-1)^2 +(y+2)^2 +2005≥2005`

nên `B` đạt GTNN là `B=2005`

khi đó \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\) `<=>`\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(A=x-x^2+\frac{1}{2}\)

\(\Leftrightarrow A=-\left(x^2-x-\frac{1}{2}\right)\)

\(\Leftrightarrow A=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{3}{4}\right)\)

\(\Leftrightarrow A=-\left[\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\right]\)

Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\)nên \(A=-\left[\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\right]\le\frac{3}{4}\)

Vậy \(A_{min}=\frac{3}{4}\)(Dấu "="\(\Leftrightarrow x=\frac{1}{2}\))

\(A_{max}=\frac{3}{4}\)nhé

D = x - x² + 3

D = - x² + x + 3

D = - ( x² - x - 3 )

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

D = - [ ( x - 1 / 2 )² - 13 / 4 ]

D = - ( x - 1 / 2 )² + 13 / 4 \(\le\)13 / 4

Dấu " = " xảy ra \(\Leftrightarrow\)x - 1 / 2 = 0

                             \(\Rightarrow\)x              = 1 / 2

Max D = 13 / 4 \(\Leftrightarrow\)x = 1 / 2

D=x-x^2+3

D= -[x^2 -x +1/4 ] + 13/4 

D=-(x-1/2)^2 +13/4 

Vì -(x-1/2)^2<=0 => D<=13/4

Dấu = xảy ra <=> x-1/2=0 <=> x=1/2