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

\(\frac{x^2+xy+y^2}{x-y}\)

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

\(=\frac{x^3-y^3}{x^2-2xy+y^2}\)

1.a) xy + 2y - x² + 4

= y ( x + 2 ) - ( x² - 4 ) = y ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 )( y - x + 2 )

b) 2x² + y² + 3xy

= ( 2x² + 2xy ) + ( y² + xy )

= 2x ( x + y ) + y ( x + y )

= ( x + y ) ( 2x + y )

2.

x - y = 5 \(\Rightarrow\)( x - y )² = 25 \(\Rightarrow\)x² + y² = 25 + 2xy = 25 + 2.3 = 31

A = ( x + y )² = x² + y² + 2xy = 31 + 6 = 37

\(a,5\left(x-y\right)-3x\left(y-x\right)=5\left(x-y\right)+3x\left(x-y\right)=\left(5+3x\right)\left(x-y\right)\\ b,x^2-4xy+4y^2=\left(x-2y\right)^2\\ c,\left(x+1\right)^2+x\left(5-x\right)=0\\ \Rightarrow x^2+2x+1+5x-x^2=0\\ \Rightarrow7x+1=0\\ \Rightarrow7x=-1\\ \Rightarrow x=-\dfrac{1}{7}\)

a: =(x-y)(5+3x)

c: \(\Leftrightarrow x^2-2x+1+5x-x^2=0\)

hay x=-1/3

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

Lời giải:
a. $(x^3+x^2y+xy^2+y^3)(x-y)=[x^2(x+y)+y^2(x+y)](x-y)$
$=(x^2+y^2)(x+y)(x-y)=(x^2+y^2)(x^2-y^2)=x^4-y^4$
b.

$(2x-1)(x+3)=2x(x+3)-(x+3)=2x^2+6x-x-3=2x^2+5x-3$

\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)

Bài 1 em dùng HĐT nha

Bài 2:

a. x² + 4x + 4

= x² + 2.2.x + 2²

= (x + 2)²

b. 9x² - 12x + 4

= (3x)² - 3x.2.2 + 2²

= (3x - 2)²

c. \(\dfrac{x^2}{4}+x+1\)

= \(\left(\dfrac{x}{2}\right)^2+2.\dfrac{x}{2}.1+1^2\)

= \(\left(\dfrac{x}{2}+1\right)^2\) 

\(\dfrac{x^2+y}{y^2+x}\)-> (x*x+y)/(y*y+x)

\(x^2+\dfrac{y}{y^2}+x\) -> x*x+y/y*y+x 

x*x và y*y có thể thay thế bằng sqr(x) và sqr(y)

Ta có:\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(y+z\right)+\left(y+z\right)^2\)

\(=\left[\left(x+y+z\right)-\left(y+z\right)\right]^2\)

\(=x^2\)

\(=x.x\)