Phân tích thành nhân tử:

(4x + 3y)² + (6xy - 2)²

=\((16x^2+24xy+9y^2)+(36x^2y^2-24xy+4)\)

=\(16x^2+24xy+9y^2+36x^2y^2-24xy+4\)

=\(16x^2+9y^2+36x^2y^2+4\)

=\((4x)^2+(3y)^2+(6xy)^2+2^2\)

MÌNH CHỈ LÀM ĐC TỚI ĐÂY

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.

1. Ta có: \(3xy\left(a^2+b^2\right)+ab\left(x^2-9y^2\right)\)

\(=3xya^2+3xyb^2+abx^2+ab9y^2\)

\(=\left(3xya^2+abx^2\right)+\left(3xyb^2+ab9y^2\right)\)

\(=ax\left(3ya+bx\right)+3by\left(xb+3ya\right)\)

\(=\left(3ya+xb\right)\left(3yb+ax\right)\)

2.Check lại đề hộ mình nha:((

Câu 2 nên sủa lại đề nha

2. xy(a²+2b²)+ab(2x²+y²)

=xya²+xy2b²+ab2x²+aby²

=(xya²+aby²)+(xy2b²+ab2x²)

=ay(ax+by)+2bx(by+ax)

=(ax+by(ay+2bx)

Ta có:  x⁶ -y⁶= (x³)²  -(y³)²  = (x³  - y³)(x³ + y³)

\(x^6-y^6\)

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

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

hk 

tốt

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)  \(4x^2-25\)

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

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

1+ x² - y² -2 =

=x² -(y+1)²

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

1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab

=>(a+b/)2ab-1/h=0

quy dong len ta co

(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0

                                                                       =>ah+bh-ab-ab=0

                                                                         =>a(h-b)-b(a-h)=0  

                                                                           =>a(h-b)=b(a-h)

                                                                              =>a/b=(a-h)(h-b)

-- là giề

a(b^3-c^3) +b(c^3-a^3)+c(a^3-b^3)

=> a(b-c)(b^2+bc+c^2)+bc^3-ba^3+ca^3-cb^3

=>a(b-c)(b^2+bc+c^2)-(cb^3-bc^3)-(ba^3-ca^3)

=>a(b-c)(b^2+bc+c^2)-bc(b-c)(b+c)-a^3(b-c)

=>(b-c)(ab^2+abc+ac^2-cb^2-bc^2-a^3)

=>(b-c)(

a(b³-c³)+b(c³-a³)+c(a³-b³)

=a.b³+b.c³+c.a³-a.c³-b.a³-c.b³

=(a.b³-b.a³)+(b.c³-c.b³)+(c.a³-c³.a)

=(ab.b²-ba.a²)+(bc.c²-bc.b²)+(ca.a²-ca.c²)

=ab(a²-b²)+bc(c²-b²)+ca(c²-a²)

có hằng đẳng thức nào ko