Câu 1: Phân tích đa thức thành nhân tử

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

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

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

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

Câu 2: Rút gọn

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

x² - 2xy + y² + 2x - 2y = (x² - 2xy + y²) + (2x - 2y)

= (x - y)² + 2(x - y)

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

Câu 2 của bn bị sai đề bài hay sao ý

Câu 2 :

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

Rồi bn triệt tiêu x+y ở tử với mẫu đi sẽ dc kết quả là \(\dfrac{x\left(x+y\right)}{x-y}\)

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

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

\(=x\left(6x+5-3y\right)\)

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

\(3x^4-48=3\left(x^4-16\right)=3\left[\left(x^2\right)^2-4^2\right]\\ =3\left(x^2-4\right)\left(x^2+4\right)\\ =3\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

\(x^3-64=x^3-4^3\)

\(\Rightarrow\left(x-4\right)\left(x^2+4x+4^2\right)\)

x³-64=x³-4³=(x-4)(x²+4x+16)

Ta có:\(x^3-64\)

\(=x^3-4^3\)

Áp dụng hằng đẳng thức:\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(\Rightarrow x^3-4^3=\left(x-4\right)\left(x^2+4x+4^2\right)\)

\(x^4+1997x^2+1996x+1997=x^4+1997x^2+1997x-x+1997=\left(x^4-x\right)+1997\left(x^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)=\left(x^2-x+1997\right)\left(x^2+x+1\right)\)

\(x^4+1997x^2+1996x+1997\\ =x^4+x^3+x^2-x^3-x^2-x+1997x^2+1997x+1997\)

\(=\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(1997x^2+1997x+1997\right)\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

k biet lam a non the