1.  \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)

\(=xya^2+2xyb^2-2abx^2-aby^2\)

\(=xya^2-aby^2-2abx^2+2xyb^2\)

\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)

\(=\left(ay-2bx\right)\left(ax-by\right)\)

2. \(xy\left(a^2+2b^2\right)+ab\left(2x^2+y^2\right)\)

\(=xya^2+2xyb^2+2abx^2+aby^2\)

\(=xya^2+aby^2+2abx^2+2xyb^2\)

\(=ay\left(ax+by\right)+2bx\left(ax+by\right)\)

\(=\left(ay+2bx\right)\left(ax+by\right)\)

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)

Mình đã làm sai, cô Hoài Thương đúng rồi.

x²+(2a+b)xy+2aby²

=x²+2axy+bxy+2aby²

=(x²+bxy)+(2axy+2aby²)

=x(x+by)+2ay(x+by)

=(x+by)(x+2ay)

cảm ơn bn

1+ x² - y² -2 =

=x² -(y+1)²

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

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

i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)

a) 1/2(x³+8)=1/2(x+2)(x²-2x+4)

b) x⁴(x-y)+2x³(x-y)=x³(x+2)(x-y)

c) x²-(y²-6y+9)=x²-(y-3)²=(x-y+3)(x+y-3)

d) xy(x³+y³)=xy(x+y)(x²-xy+y²)

e)3x²(x²-25y²)=3x²(x-5y)(x+5y)

f) 4x⁴+4x²y²+y⁴-4x²y²= (2x²+y²)²-(2xy)²=(2x²-2xy+y²)(2x²+2xy+y²)

a) \(\frac{1}{2}x^3+4=\frac{1}{2}\left(x^3+8\right)=\frac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)

b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)

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

d) \(x^4y+xy^4=xy\left(x^3+y^3\right)=xy\left(x+y\right)\left(x^2-xy+y^2\right)\)

e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).

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

mình k ghi lại đề nha bạn

\(=\left(x-y\right)^2-16z^2\\ =\left(x-y-4z\right)\left(x-y+4z\right)\)

x²-6xy+9y²-16z²

=[x²-2.3xy+(3y)²]-16z²

=[x-3y]²-[4y]²

=[x-3y-3z][x-3y+3z]

Lời giải trước mk đánh nhầm nha

x²-6xy+9y²-16z²

=[x²-2.3xy+(3y)²]-16z²

=[x-3y]²-[4z]²

=[x-3y-4z][x-3y+4z]

a) \(\left(a+b\right)^3+\left(a+b\right)^3\)

\(=\left(a+b+a+b\right)\left[\left(a+b\right)^2-2\left(a+b\right)^2+\left(a+b\right)^2\right]\)

\(=2\left(a+b\right)\left[\left(a+b\right)^2\left(1-2+1\right)\right]\)

\(=2\left(a+b\right)\)

b)  \(9x^2+6xy+y^2\)

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

\(=\left(3x+y\right)\left(3x+y\right)\)

c)  \(4x^2-25\)

\(=\left(2x\right)^2-5^2\)

\(=\left(2x+5\right)\left(2x-5\right)\)