a,<=>   x²-4x+2²+y²-8y+4²-14

<=> (x²-2x2+2²)+(y²-2x4+4²)-14

<=> (x-2)²+(y-4)²-14 

Vì (x-2)²+(y-4)²>= 0

=> F >= -14 => MIn F = -14 <=> x=2, y=4

b, <=> (x²+5²+(2y)²-4xy+10x-20y) +(y²-2y+1)+2

<=> (x+5-2y )²+(y-1)²+2 

Vì (x+5-2y) ²+(y-1)² >= 0

=> G >= 2 => Min =2 <=> y=1, x= -3

\(F=x^2-4x+y^2-8y+6\)

\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)

\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

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

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

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)

\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)

c: Để A=3/4 thì 4x-8=3x+6

=>x=14

d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{3;1;4;0;6;-2\right\}\)

a: \(A=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{1}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu '=' xảy ra khi x=5/2

b: \(B=x^2-4x+4+y^2-8y+16-14\)

\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

Dấu '=' xảy ra khi x=2 và y=4

a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)

\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)

hay P=x+3

<=> xaa ) C= x²-6x + 11= (x-3)² +2

ta co : (x-3)² + > hoặc = 2

=> C đạt giá trị nhỏ nhất khi C=2

<=> x=3

b) D =(x-1) (x+2)(x+3)(x+6)

= [ (x-1)(x+6)][(x+2)(x+3)]

=(x² +5x -6)(x²+5x +6)

=(x²+5x )² - 36

ta có (x² +5x)² -36 luôn > hoặc = -36

=> D đạt GTNN khi D = -36

<=>(x² + 5x)^{2 =0}

=> x = 0 hoac x =-5

c) E = x² - 4x + y² - 8y + 6

=(x² -4x +4 ) + (y² - 8y +16 ) -14

= (x -2)² +( y-4)² -14

ta co (x-2)² + (y-4)² -14 luôn > hoặc = -14

=> E dat GTNN khi E = -14

<=> (x-2)²​ =0 va (y-4)² =0

<=> x =2 va y=4

d) G =x² -4xy +5y² + 10x -22y + 28 ( de sai nha ban )

= [(x² - 4xy + 4y² ) + 10x -20y +25 ]+ ( y² -2y +1 ) +2

= [(x-2y)² + 10x - 20y + 25 ] + (y-1)² +2

= [( x-2y)² + 2. 5 (x-2y) + 25 ] + (y-1)² +2

= (x-2y +5)² + ( y-1)² +2

ta co (x-2y +5 )² + (y-1)² +2 luôn > hoặc = 0

=> G đạt GTNN khi (x-2y+5 )²=0 hoac (y-1)² =0

<=> y-1 = 0 => y = 1

,=> x =-3

- thanks bạn :v

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

\(\Leftrightarrow\left(x-1\right)^2>=0\)

hay \(x\in R\)