x³ + 2x² - 3x = x³ + 3x² - x² - 3x = x². (x +3) - x(x+3) = (x² - x).(x+3)

=> (ax² + bx + c).(x + 3) = (x² - x)(x + 3)

=>ax²  + bx + c = x² - x với mọi x

=> a = 1; b = -1; c = 0 

\(=x^4-8x^3+24x^2-32x+16+x^4+8x^3+24x^2+32x+16=2x^4+48x^2+32=2\left(x^4+24x+16\right)\)

đây là 1 hằng đẳng thức luôn

\(=\left(2x-3-2x-5\right)^2=\left(-8\right)^2=64\)

\(A=\left(4x^2-2.\frac{1}{2}2.x+\frac{1}{4}\right)+\frac{47}{4}=\left(2x+\frac{1}{2}\right)^2+\frac{47}{4}\ge\frac{47}{4}\Rightarrow MinA=\frac{47}{4}\Leftrightarrow x=-\frac{1}{4}\)

a) \(=x^2\left(a-b\right)-2\left(a-b\right)=\left(a-b\right)\left(x^2-2\right)\)

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

c) \(50x^2\left(x-y\right)^2-8y^2\left(x-y\right)^2=\left(x-y\right)^2\left(50x^2-8y^2\right)\)

d) \(15a^mb.15^2-15a^mb.3=15^mb\left(15^2-3\right)=15^mb.222\)

= (x + y)³ + z³ – 3x2y – 3xy2 - 3xyz 

= (x + y +z)[(x + y)² – (x + y)z + z²)] - 3xy(x + y + z)

= (x + y + z)(x² +2xy + y² – xz – yz +z² – 3xy)

= (x + y + z)(x² + y² +z² – xy - yz – xz)

x³ - y³ - z³ +3xyz

= (x³ - 3x²y +3xy² -y^3) +3x²y-3xy² - z³ +3xyz 

= [(x-y)³ -z³] + 3x²y -3xy² +3xyz

= (x-y-z)(x² + 2xy+y² +zx+zy + z²) + 3xy( x-y+z)

x3−y3−z3+3xyz=(x+y+z)(xy+yz+xz−x2−y2−z2) =-(x^3+y^3+y^3-3xyz)$

Ta tính x3+y2+z3−3xyz trước

ta có:

x3+y3+z3−3xyz=(x+y)3+z3−3xy(x+y)−3xyz=(x+y+z)(x2+y2+z2−xy−yz−xz)

=>x3−y3−z3+3xyz=(x+y+z)(xy+zy+xz−x2−y2−z2)