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\(A=x-x^2+\frac{1}{2}\)
\(\Leftrightarrow A=-\left(x^2-x-\frac{1}{2}\right)\)
\(\Leftrightarrow A=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{3}{4}\right)\)
\(\Leftrightarrow A=-\left[\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\right]\)
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 "="\(\Leftrightarrow x=\frac{1}{2}\))
(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....
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
`a)A=-x^2+x+1`
`=-(x^2-x)+1`
`=-(x^2-2.x. 1/2+1/4-1/4)+1`
`=-(x-1/2)^2+5/4<=5/4`
Dấu "=" xảy ra khi `x-1/2=0<=>x=1/2`
`b)B=x^2+3x+4`
`=x^2+2.x. 3/2+9/4+7/4`
`=(x-3/2)^2+7/4>=7/4`
Dấu "=" xảy ra khi `x-3/2=0<=>x=3/2`
`c)=x^2-11x+30`
`=x^2-2.x. 11/2+121/4-1/4`
`=(x-11/2)^2-1/4>=-1/4`
Dấu "=" xảy ra khi `x+1/4=0<=>x=-1/4`
N = x2 + x + 1
= x2 + 2.x.\(\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
= \(\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\)với mọi x
\(\Rightarrow\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)
hay \(N\ge\frac{-1}{4}\)
Dấu " = " xảy ra <=> \(x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
Vậy GTNN của \(N=\frac{-1}{4}\Leftrightarrow x=\frac{-1}{2}\)
B=y^2-y+1
=y^2-2*y*1/2+1/4+3/4
=(y-1/2)^2+3/4>=3/4
Dấu = xảy ra khi y=1/2
E=-x^2+x+2
=-(x^2-x-2)
=-(x^2-x+1/4-9/4)
=-(x-1/2)^2+9/4<=9/4
Dấu = xảy ra khi x=1/2
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
\(A=4-x^2+3\)
\(=-x^2+7\le7\)
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
\(=t\left(t+2\right)=t^2+2t+1-1\)
\(=\left(t+1\right)^2-1\ge-1\)
Đặt \(A=x^2+x+1\)
\(=x^2+2\cdot\frac{1}{2}\cdot x+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu " = " xảy ra khi và chỉ khi
\(\left(x+\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
vậy \(A_{min}=\frac{3}{4}\) tại \(x=-\frac{1}{2}\)
\(x^2+x+1=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge\left(\forall x\right)\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
Vậy GTNN của x2 + x + 1 bằng 3/4 khi và chỉ khi x = -1/2