`x^2 - 4x+1`

`= x^2 -2.x.2+2^2 -3`

`= (x-2)^2 -3 >= -3` với mọi `x`

Dấu "=" xảy ra khi :

`(x-2)^2=0 <=>x-2=0 <=>x=2`

Vậy GTNN của BT là `-3 <=>x=2`

Bài 10: c) 

\(C=8x^3-27y^3=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x-3y\right)\left[\left(2x-3y\right)^2+18xy\right]\)

\(=5.\left(5^2+18.4\right)=485\)

Bài 11: 

\(\left(a+b+c\right)^3=\left(a+b\right)^3+c^3+3\left(a+b\right).c.\left(a+b+c\right)\)

\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

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

Bài 12: 

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)^3-3c\left(a+b\right)\left(a+b+c\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ab-3bc-3ca\right]\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

`a,`

`(x+y)^3-1=(x+y)^3-1^3=(x+y-1)[(x+y)^2 +x+y +1]  =(x+y-1)(x^2 +2xy+y^2 +x+y+1]`

`b,`

`100x^2 - (x^2 +25)^2=(10x)^2-(x^2 +25)^2=(10x-x^2-25)(10x +x^2 +25) = -(x-5)^2 (x+5)^2`

a) \(\left(x+y\right)^3-1\)

\(=\left(x+y\right)^3-1^3\)

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

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

b) \(100x^2-\left(x^2+25\right)^2\)

\(=\left(10x\right)^2-\left(x^2+25\right)^2\)

\(=[10x-\left(x^2+25\right)][10x+\left(x^2+25\right)]\)

\(=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)

\(=\left(-x^2+10x-25\right)\left(x^2+10x+25\right)\)

\(=-\left(x^2-10x+25\right)\left(x^2+10x+25\right)\)

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