\(A=2a^2+b^2-2ab+10a+42=\left(a^2-2ab+b^2\right)+\left(a^2+10a+25\right)+17=\left(a-b\right)^2+\left(a+5\right)^2+17\ge17\)

\(minA=17\Leftrightarrow a=b=-5\)

C=2a²+b²-2ab+10a+42

=a²-2ab+b²+a²+10a+25+17

=(a-b)²+(a+5)²+17

=>MIN(C)=17 <=>a-b=0 và a+5=0

<=>a=b=-5

vậy ..................

$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$

$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
Vậy $A_{\min}=7$. Giá trị này đạt tại $(x-3)^2=(y+2)^2=0$

$\Leftrightarrow x=3; y=-2$

---------------------

$B=9x^2+y^2+2z^2-18x+4z-6y+30$

$=(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)+10$

$=9(x^2-2x+1)+(y^2-6y+9)+2(z^2+2z+1)+10$

$=9(x-1)^2+(y-3)^2+2(z+1)^2+10\geq 10$
Vậy $B_{\min}=10$. Giá trị này đạt tại $(x-1)^2=(y-3)^2=(z+1)^2$

$\Leftrightarrow x=1; y=3; z=-1$

$C=x^2+y^2+z^2-xy-yz-xz+3$

$2C=2x^2+2y^2+2z^2-2xy-2yz-2xz+6$

$=(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)+6$

$=(x-y)^2+(y-z)^2+(z-x)^2+6\geq 6$

$\Rightarrow C\geq 3$

Vậy $C_{\min}=3$. Giá trị này đạt tại $x-y=y-z=z-x=0$

$\Leftrihgtarrow x=y=z$

--------------------------------------

$D=5x^2+2y^2+4xy-2x+4y+2021$

$=2(y^2+2xy+x^2)+3x^2-2x+4y+2021$

$=2(x+y)^2+4(x+y)+3x^2-6x+2021$
$=2(x+y)^2+4(x+y)+2+3(x^2-2x+1)+2016$

$=2[(x+y)^2+2(x+y)+1]+3(x^2-2x+1)+2016$

$=2(x+y+1)^2+3(x-1)^2+2016\geq 2016$

Vậy $D_{\min}=2016$ khi $x+y+1=x-1=0$

$\Leftrightarrow x=1; y=-2$

$E=x^2-2x+4y^2+4y+2014$

$=(x^2-2x+1)+(4y^2+4y+1)+2012$

$=(x-1)^2+(2y+1)^2+2012$

$\geq 2012$

Vậy $E_{\min}=2012$. Giá trị này đạt tại $x-1=2y+1=0$

$\Leftrightarrow x=1; y=\frac{-1}{2}$

----------------------

$F=5x^2+5y^2+8xy+2y-2x+30$

$=4(x^2+2xy+y^2)+x^2+y^2+2y-2x+30$

$=4(x+y)^2+(x^2-2x+1)+(y^2+2y+1)+28$

$=4(x+y)^2+(x-1)^2+(y+1)^2+28\geq 28$

Vậy $F_{\min}=28$. Giá trị này đạt tại $x+y=x-1=y+1=0$

$\Leftrightarrow x=1; y=-1$

1) \(a^2-2ab+b^2-9=\left(a-b\right)^2-9=\left(a-b-3\right)\left(a-b+3\right)\)

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

3) \(25-x^2-2xy-y^2=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)

\(1,=\left(a-b\right)^2-9=\left(a-b-3\right)\left(a-b+3\right)\\ 2,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 3,=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)

\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{1}{a+b}\)

\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{1}{a-b}\)

\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)

a,x²-y²-2x+2y
= (x+y)(x-y) - 2(x-y)
= (x-y)(x+y-2)
b,2x+2y-x²-xy
= 2(x+y) - x(x+y)
= (x+y)(2-x)
c,3a²-6ab+3b²-12c²
= 3(a² - 2ab + b² - 4c²)
= 3[(a-b)² - 4c²)
= 3(a-b-2c)(a-b+2c)
d,x²-25+y²+2xy
= (x+y)² - 25
= (x+y+5)(x+y-5)

e) a²+2ab+b²-ac-bc

= (a+b)²-c(a+b)

= (a+b)( a+b-c)

f) x²-2x-4x²-4y

= -3x²-2x-4y

= -(3x²+2x+4y)

g)x²y-x³-9y+9x

= x²(y-x)-9(y-x)

= (y-x)(x²-9)

h) x²(x-1)+16(1-x)

= x²(x-1)-16(x-1)

= (x-1)(x²-16)

= (x-1)(x-4)(x+4)

n) 81x²-6yz-9y²-z²

= (9x)²-[(3y)²+6yz+z²]

=(9x)²-(3y+z)²

=(9x+3y+z)(9x-3y-z)

m) xz- yz-x²+2xy-y²

= z(x-y)-(x²-2xy+y²)

= z(x-y)-(x-y)²

= (x-y)(z-x+y)

 p) x² + 8x + 15

= x² + 3x + 5x + 15

= x(x+3) + 5(x+3)

= (x+3)(x+5)

k) x² - x - 12

= x² + 3x - 4x - 12

= x(x+3) - 4(x+3)

= (x+3)(x-4)

\(a,=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4-x-x^2+x^3-x^4+x^5+1+x-x^2+x^3-x^4\\ =2x-2x^2+2x^3-2x^4\)

Bài 1:

$A=(9x^2-5x)+(5y^2+3y)$

$=[(3x)^2-2.3x.\frac{5}{6}+(\frac{5}{6})^2]+5(y^2+\frac{3}{5}y+\frac{3^2}{10^2})-\frac{103}{90}$

$=(3x-\frac{5}{6})^2+5(y+\frac{3}{10})^2-\frac{103}{90}$

$\geq \frac{-103}{90}$

Vậy $A_{\min}=\frac{-103}{90}$. Giá trị này đạt tại $3x-\frac{5}{6}=y+\frac{3}{10}=0$

$\Leftrightarrow (x,y)=(\frac{5}{18}, \frac{-3}{10})$

Bài 2:

a. 

$-A=4x^2+5y^2-8xy-10y-12$

$=(4x^2-8xy+4y^2)+(y^2-10y+25)-37$

$=(2x-2y)^2+(y-5)^2-37\geq -37$

$\Rightarrow A\leq 37$

Vậy $A_{\max}=37$. Giá trị này đạt tại $2x-2y=y-5=0$

$\Leftrightarrow x=y=5$

b.

$-B=3x^2+16y^2+8xy+5x-2$

$=(x^2+16y^2+8xy)+2(x^2+\frac{5}{2}x+\frac{5^2}{4^2})-\frac{41}{8}$

$=(x+4y)^2+2(x+\frac{5}{4})^2-\frac{41}{8}$

$\geq \frac{-41}{8}$

$\Rightarrow B\leq \frac{41}{8}$
Vậy $B_{\max}=\frac{41}{8}$. Giá trị này đạt tại $x+4y=x+\frac{5}{4}=0$

$\Leftrightarrow x=\frac{-5}{4}; y=\frac{5}{16}$

Bài 3: 

a: Ta có: \(\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\)

\(=y^2+8y-5y-40-y^2+y-4y+4\)

=-36

b: Ta có: \(y^4-\left(y^2-1\right)\left(y^2+1\right)\)

\(=y^4-y^4+1\)

=1

Bài 2: 

a: \(\left(2a-b\right)\left(4a+b\right)+2a\left(b-3a\right)\)

\(=8a^2+2ab-4ab-b^2+2ab-6a^2\)

\(=2a^2-b^2\)

b: \(\left(3a-2b\right)\left(2a-3b\right)-6a\left(a-b\right)\)

\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)

\(=6b^2-7ab\)

c: \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)

\(=10bx-5b^2-16bx+8b^2+2x^2-xb\)

\(=3b^2-7xb+2x^2\)