1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

(x^2-x+2)^2+(x-2)^2 
= [(x^2-x+2)+(x-2)]^2-2[(x^2-x+2)*(x-2)] (áp dụng (a^2+b^2)=(a+b)^2-2ab 
=(x^2)^2- 2((x^3-3x^2+4x-4) 
=x^4-2x^3+6x^2-8x+8 
 giờ phân tích đa thức 
x^4-2x^3+6x^2+8x-8 
=(x^4-2x^3+2x^2)+(4x^2-8x+8) (cái này làm bài tập nhiêu nhìn ra nhanh) 
=[x^2(x^2-2x+2)]+4(x^2-2x+2) dẹp luôn 
=(x^2-2x+2)(x^2+4) 

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

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

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

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

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

Cảm ơn hai bạn nha nhưng Đường Quỳnh Giang nhầm dấu rồi.😆

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

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

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

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

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

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

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

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.

\(8x^3+12x^2y+6xy^2+y^3-z^3\)

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

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

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2

x^3-x+6=x^3+2x^2-2x^2-4x+3x+6=x^2.(x+2)-2x.(x+2)+3.(x+2)=(x^2-2x+3).(x+2)

\(-2x^3+x^2+12\)

\(=-2x^3+4x^2-3x^2+6x-6x+12\)

\(=-2x^2\left(x-2\right)-3x\left(x-2\right)-6\left(x-2\right)\)

\(=\left(x-2\right)\left(-2x^2-3x-6\right)\)

\(8x^4+81\)

\(=8x^4+2\cdot2\sqrt{2}\cdot x^2\cdot9+81-36\sqrt{2}\cdot x^2\)

\(=\left(2\sqrt{2}x^2+9\right)^2-\left(6\sqrt[4]{2}\cdot x\right)^2\)

\(=\left(2\sqrt{2}\cdot x^2-6\sqrt[4]{2}\cdot x+9\right)\left(2\sqrt{2}\cdot x^2+6\sqrt[4]{2}\cdot x+9\right)\)