60 số ( đúng thì k, sai thì thôi :D)

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

\(x^6-y^6\)

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

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

\(x^2-4y^2+16x+64=\left(x^2+16x+64\right)-4y^2\)

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

\(=\left(x-2y+8\right)\left(x+2y+8\right)\)

a) \(\frac{9x^3y^2}{3xy^3}=\frac{3x^2}{y}\)

b) \(\frac{4xy\left(x+y\right)}{6xy^2\left(x+y\right)^2}=\frac{2}{3y\left(x+y\right)}\)

c) \(\frac{3x^2+3x}{x+1}=\frac{3x\left(x+1\right)}{x+1}=3x\)

d)  \(\frac{3x^2+6x}{x^2\left(x+2\right)}=\frac{3x\left(x+2\right)}{x^2\left(x+2\right)}=\frac{3}{x}\)

\(6x^2y:2xy\)

\(=2xy.3x:2xy\)

\(=3x\)

\(H=\left(x^2-2x-3\right)\left(x^2-2x+1\right)=\left(t-3\right)\left(t+1\right)=t^2-2t-3=\left(t-1\right)^2-4\ge-4\)

Với t = x² - 2x

Vậy GTNN của H là -4

a. Ta có : (a+b+c)² ≤ 3(a²+b²+c²)

⇌ a²+b²+c²+2ab+2ac+2bc ≤ 3a²+3b²+3c²

⇌ -2a²-2b²-2c²+2ab+2ac+2bc ≤ 0

⇌ -(a-b)²-(b-c)²-(a-c)² ≤ 0 (đúng với mọi a,b,c)

Dấu "=" xảy ra khi a=b=c

b. Ta có: (a+b)² ≤ 2(a²+b²)

⇔ a²-2ab+b² ≤ 2a²+2b²

⇔ -a²-2ab-b² ≤ 0

⇔ -(a+b)² ≤ 0 (đúng với mọi a,b)

Dấu "=" xảy a khi a=-b