\(\dfrac{1}{2}x^3+4\)

\(=\dfrac{x^2+8}{2}\)

\(=\text{(x+2)(x^2 −2x+4)}\)

\(R=-2\left(x^2-\dfrac{1}{2}x-\dfrac{1}{2}\right)=-2\left(x^2-2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{9}{16}\right)\)

\(=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\forall x\)

Dấu '=' xảy ra khi x=1/4

Ta có: M=−x2−2x+5

=−(x2+2x−5)

=−(x2+2x+1)+6

=−(x+1)2+6

Vì −(x+1)2≤0∀x

⇒−(x+1)2+6≤6∀x

Dấu "=" xảy ra ⇔

$x=-1$

Vậy $MA{X}_M=6\iff x=-1$

Đặt A=4x−x2+3

=−x2+4x+3=−(x2−4x−3)

=−(x2−4x+4−7)

=−[(x−2)2−7]

=−(x−2)2+7

Ta có: −(x−2)2≤0⇒−(x−2)2+7≤7

Dấu " = " khi (x−2)2=0⇔x=2

Vậy MAXA=7 khi x = 2

\(A=2x-2x^2-5\)

\(A=-2\left(x^2-x\right)-5\)

\(A=-2\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{2}-5\)

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

Có \(2\left(x-\frac{1}{2}\right)^2\ge0\)với mọi x

=> \(-2\left(x-\frac{1}{2}\right)^2\le0\)với mọi x

=> \(-2\left(x-\frac{1}{2}\right)^2-4\frac{1}{2}\le-4\frac{1}{2}\)với mọi x

=> \(A\le-4\frac{1}{2}\)với mọi x

Dấu "=" xảy ra <=> \(x-\frac{1}{2}=0\)<=> \(x=\frac{1}{2}\)

KL: \(A_{max}=-4\frac{1}{2}\)<=> \(x=\frac{1}{2}\)

2x - 2x² - 5

= -2( x² - x + 1/4 ) - 9/2

= -2( x - 1/2 )² - 9/2 ≤ -9/2 ∀ x

Dấu "=" xảy ra <=> x = 1/2

Vậy GTLN của biểu thức = -9/2 <=> x = 1/2

a) 3xy² - 3x³ - 6xy + 3x 

=3x (y² - x² - 2y +1)

= 3x [ (y-1)² -x² ]

=3x (y-1-x)(y-1+x)

b) 3x² +11x+6

= 3 x² +9x +2x +6

=3x (x+3)+2(x+3)

= (x+3)(3x+2)

a. \(A=4x-x^2+3=7-\left(x^2-4x\right)+4=7-\left(x-2\right)^2\le7\)

b.\(B=x-x^2=\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)

c.\(C=2x-2x^2-5=-\frac{9}{2}-2\left(x^2-x+\frac{1}{4}\right)=-\frac{9}{2}-2\left(x-\frac{1}{2}\right)^2\le-\frac{9}{2}\)

giải câu b trc nha

= ((x-1)^2+2009]/x^2=(x-1)^2/x^2+2009

vậy min=2009 khi x=1

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