\(x^2\left(x-1\right)+16\left(1-x\right)\)

\(=x^2\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-16\right)\)

\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

\(P=x^2-6xy+9y^2=\left(x-3y\right)^2\)

(Áp dụng 7 hằng đẳng thức đáng nhớ)

c/ 3x-9xy-9y²+6y-1

=3x.(1-3y)-(1-6y+9y²)

=3x.(1-3y)-(1-3y)²

=(1-3y)[3x-(1-3y)]

=(1-3y)(3x-1+3y)

d/ x⁴+x²y²+y⁴

=x⁴+2x²y²+y⁴-x²y²

=(x²+y²)²-x²y²

=(x²-xy+y²)(x²+xy+y²)

\(x^2-6xy+9y^2=x^2-2.\left[x.\left(3y\right)\right]+\left(3y\right)^2\)

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

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)

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

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

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

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

a) \(-y^2+\dfrac{1}{9}\)

\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)

\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)

b) \(4^4-256\)

\(=4^4-4^4\)

\(=0\)

c) \(9\left(x-3\right)^2-4\left(x+1\right)^2\)

\(=\left(3x-9\right)^2-\left(2x+2\right)^2\)

\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)

\(=\left(5x-7\right)\left(x-11\right)\)

\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)

\(49.\left(y-4\right)^2-9y^2-36y-36\)

\(=7^2\left(y-4\right)^2-\left(9y^2+36y+36\right)\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28+3y+6\right).\left(7y-28-3y-6\right)\)

\(=\left(10y-22\right).\left(4y-34\right)\)

\(=4.\left(5y-11\right).\left(2y-17\right)\)

\(49\left(y-4\right)^2-9y^2-36y-36\)

\(=\) \(4\left(2y-17\right)\left(5y-11\right)\)

x^3-2x^2-x+2=x^2(x-2)-(x-2)=(x^2-1)(x-2)=(x-1)(x+1)(x-2)

a)(x³- x) - (2x² - 2)

= x (x² - 1) - 2 (x² - 1)

= (x - 2) (x²-1)

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

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

b) \(x^2+6x-y^2+9=\left(x^2+6x+9\right)-y^2\)

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

y² hay y ạ???