a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)

\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)

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

\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)

c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)

\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)

a) \(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)

b) \(=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)

c) \(=3x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(3x+5\right)\)

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

e) \(=x\left(x^2-11x+30\right)\)

f) \(=x\left(x-3\right)+6\left(x-3\right)=\left(x-3\right)\left(x+6\right)\)

A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

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

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

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

a) A = a) A = x² - xy + x - y= (x² - xy) + (x - y)=x(x-y)+(x-y)=(x+1)(x-y)
c) A = 3x - 3y + x² - y²=3(x-y)+(x-y)(x+y)=(3+x+y)(x-y)
d) A = x² - y² - 2x - 2y = (x-y)(x+y)-2(x+y)=(x+y)(x-y-2)

câu b bạn xem lại đúng đề ko
 

\(\)a, \(A=x^2-xy+x-y\)
        \(=x\left(x-y\right)+\left(x-y\right)\)
        \(=\left(x+1\right)\left(x-y\right)\)
 

a) Ta có: ( x2 -1 )( x2 + 2x )

= x2( x2 + 2x ) - ( x2 + 2x )

= x4 + 2x3 - x2 - 2x

b) Ta có ( x + 3 )( x2 + 3x -5 )

= x( x2 + 3x -5 ) + 3( x2 + 3x -5 )

= x3 + 3x2 - 5x + 3x2 + 9x - 15

= x3 + 6x2 + 4x - 15

c) Ta có ( x -2y )( x2y2 - xy + 2y )

= x( x2y2 - xy + 2y ) - 2y( x2y2 - xy + 2y )

= x3y2 - x2y + 2xy - 2x2y3 + 2xy2 - 4y2

d) Ta có ( 1/2xy -1 )( x3 -2x -6 )

= 1/2xy( x3 -2x -6 ) - ( x3 -2x -6 )

= 1/2x4y - x2y - 3xy - x3 + 2x + 6

\(A=x^2+y^2-xy+2x-2y+2022\)

\(A=\left(x^2+\dfrac{y^2}{4}+1-xy+2x-y\right)+\dfrac{3}{4}y^2-y+2021\)

\(A=\left(x-\dfrac{y}{2}+1\right)^2+\dfrac{3}{4}\left(y-\dfrac{2}{3}\right)^2+\dfrac{6062}{3}\ge\dfrac{6062}{3}\)

\(A_{min}=\dfrac{6062}{3}\) khi \(\left(x;y\right)=\left(-\dfrac{2}{3};\dfrac{2}{3}\right)\)

[Xx2+2x2]-[2-yx2-x]=6

Xx[8+1]-y=6

X:9-y=6

xy=6x9

xy=54

=>x=5

y=4

ông đi qua bà đi lại cho con xin 1 cái tik

x2-xy-2x-2y=x2-x2-xy-2y=xy-2y=y(x-2)=6

    mà 6=2x3=3x2=1x6=6x1=-2 x -3=-3 x -2= -1 x-6= -6 x -1

từ đó lập bảng rồi tính bạn ạ ;)

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)