a)

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

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

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

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

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

Ta có:

\(\left(x-\frac{1}{2}\right)^2\ge0;\left|3x+2y\right|\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\left|3x+2y\right|\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left|3x+2y\right|+2006\ge2006\)

Dấu "=" xảy ra tại \(\hept{\begin{cases}x-\frac{1}{2}=0\\3x=-2y\end{cases}}\Rightarrow x=\frac{1}{2};y=-\frac{3}{4}\)

Vậy \(A_{min}=2006\Leftrightarrow x=\frac{1}{2};y=-\frac{3}{4}\)

GTNN:

\(\Leftrightarrow x^2+2\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}+1\)

\(\Leftrightarrow x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy Min của biểu thức trên =3/4 khi x+1/2=0 => x=-1/2

GTLL:

\(\Leftrightarrow-3\left(x^2-\frac{7}{3}x-\frac{1}{3}\right)\)

\(\Leftrightarrow-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{49}{36}-\frac{1}{3}\right)\)

\(\Leftrightarrow-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{61}{36}\right)\)

\(\Leftrightarrow-3\left[\left(x-\frac{7}{6}\right)^2-\frac{61}{36}\right]\)

\(\Leftrightarrow-3\left(x-\frac{7}{6}\right)^2+\frac{61}{12}\le\frac{61}{12}\)

Vậy Max của biểu thức trên = 61/12 khi x-7/6=0 => x=7/6

nha . cảm ơn . chúc bạn học tốt

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4

a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)

Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)

\(\Rightarrow\left(2x-3\right)^2+91\ge91\)

hay A \(\ge91\)

Dấu "=" xảy ra <=> \(\left(2x-3\right)^2=0\)

<=> 2x-3=0

<=> 2x=3

<=> \(x=\frac{3}{2}\)

Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)

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

Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)

Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)

\(C=2x^2+2xy+y^2-2x+2y+2\)

\(C=x^2+2x\left(y-1\right)+\left(y-1\right)^2+x^2+1\)

\(\Leftrightarrow C=\left(x+y-1\right)^2+x^2+1\)

Ta có: 

\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)

\(\Leftrightarrow\left(x+y-1\right)^2+x^2+1\ge1\)

hay C\(\ge\)1

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+y-1\right)^2=0\\x^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=1\\x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=0\end{cases}}}\)

Vậy Min C=1 đạt được khi y=1 và x=0

Ta có : A = x² + 3x + 3 

=> A = x² + 3x + \(\frac{9}{4}+\frac{3}{4}\)

\(\Rightarrow A=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\) \(\ge\frac{3}{4}\forall x\in R\)

Vậy A_min = \(\frac{3}{4}\) khi \(x=-\frac{3}{2}\)

Từ gt ta có x^2+y^^2=xy+1

=>P=(x^2+y^2)^2-2x^2y^2-x^2y^2

=(xy+1)²-2x²y²-x²y²

=x²y²+xy+1-3x²y²=-2x²y²+xy+1

=......

\(1=x^2+y^2-xy\ge2xy-xy=xy\Rightarrow xy\le1\)

\(1=x^2+y^2-xy\ge-2xy-xy=-3xy\Rightarrow xy\ge-\dfrac{1}{3}\)

\(\Rightarrow-\dfrac{1}{3}\le xy\le1\)

\(P=\left(x^2+y^2\right)^2-2\left(xy\right)^2-\left(xy\right)^2=\left(xy+1\right)^2-3\left(xy\right)^2=-2\left(xy\right)^2+2xy+1\)

Đặt \(xy=t\in\left[-\dfrac{1}{3};1\right]\)

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

\(f'\left(t\right)=-4t+2=0\Rightarrow t=\dfrac{1}{2}\)

\(f\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{3}{2}\) ; \(f\left(1\right)=1\)

\(\Rightarrow P_{max}=\dfrac{3}{2}\) ; \(P_{min}=\dfrac{1}{9}\)

Áp dụng BĐT  Bunyakovsky  ta có:

    \(\left(x+2y\right)^2=\left(x+\sqrt{2}.\sqrt{2}y\right)^2\le\left(1^2+\sqrt{2}^2\right)\left[x^2+\left(\sqrt{2}y\right)^2\right]\)

\(\Leftrightarrow\)\(\left(x+2y\right)^2\le3\left(x^2+2y^2\right)\)

\(\Leftrightarrow\)\(1\le3\left(x^2+2y^2\right)\) (do  x + 2y = 1 )

\(\Leftrightarrow\)\(x^2+2y^2\ge\frac{1}{3}\)

Dấu "="  xảy ra  \(\Leftrightarrow\)\(\hept{\begin{cases}x+2y=1\\\frac{1}{x}=\frac{\sqrt{2}}{\sqrt{2}y}\end{cases}}\)\(\Leftrightarrow\)\(x=y=\frac{1}{3}\)

Vậy  \(Min\)\(A=\frac{1}{3}\) \(\Leftrightarrow\)\(x=y=\frac{1}{3}\)

P/s: tham khảo thôi nhé, mk ko chắc đúng (yếu phần cực trị)

\(x^2+2y^2=\left(x+2y\right)^2\) mà \(x+2y=1=>\left(x+2y\right)^2=1^2=1\)

vậy A=1