(x²-2x+1)-(4y²-4y+1)

(x-1)²- (2y-1)²

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

tích đúng cho mình nhé bn ơi

ta có :  \(x^2-2x+y^2-4y+5\)

           \(=\left(x^2-2.x.1+1\right)+\left(y^2-2.2.y+4\right)\)

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

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

       đến đấy mk  ..........

Đến đấy rs làm sao nữa

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

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

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

a) x⁴ + 4y²=(x²)²+(2y)²=(x²)²+4x²y²+(2y)²-4x²y²=(x²+y²)²-(2xy)²=(x²+y²-2xy)(x²+y²+2xy)

b) x^5+x+1=x⁵−x⁴+x²+x⁴−x²+x+x³−x²+1=(x⁵−x⁴​+x²)+(x⁴−x²+x)+(x³−x²+1)

= x²(x³-x²+1)+x(x³-x+1)+(x³−x²+1)

= (x²+x+1)(x³+x²+1)

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)