x² - 5x + 5

= x² - 5(x + 1)

= x² - \(\left[\sqrt{5\left(x+1\right)}\right]^2\)

= [x - \(\sqrt{5\left(x+1\right)}\)][x + \(\sqrt{5\left(x+1\right)}\)]

\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)

x(x-4)+5(x-4)

(x+5)(x-4)

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

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

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

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

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

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

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

\(=\left(y-1\right)\left(1-x\right)\left[x+y\left(1+x\right)\right]=\left(y-1\right)\left(1-x\right)\left(xy+x+y\right)\)

Ta có:

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

\(\Rightarrow5x^2+3.\left(x^2+2xy+y^2\right)-5y^2\)

\(\Rightarrow5x^2+3x^2+6xy+3y^2-5y^2\)

\(\Rightarrow8x^2+6xy-2y^2\)

Ta có :\(5x^2+3\left(x+y\right)^2-5y^2\)

\(\Rightarrow5x^2+3.\left(x^2-2xy-y^2\right)-5y^2\)

\(\Rightarrow5x^2+3x^2+6xy+3y^2-5y^2\)

\(\Rightarrow8x^2+6xy-2y^2\)

TK CHO MK NHA

Chúc Bạn Học Tốt :):):):)

sai het rui

x^2-y^2-(x+y)

(x-y).(x+y)-(x-y).1

(x-y).(x+y-1)

=(x²-y²)-(x+y)

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

=(x+y)(x-y-1)

\(-x-y^2+x^2-y\)

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

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

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

x² - 3x + 3y - y²

= (x² - y²) - (3x - 3y)

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

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

= x² - y² - 3x+3y = (x-y)(x+y) -3(x-y)

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

nhớ chọn cho mk nha!!!!!!