a) 15x² + 10x = 5x ( 3x +2 )

b) x² - y² +2x -2y

= (x-y) (x+y) + 2(x-y)

= (x - y) ( x + y + 2)

c) 3x² + 6xy + 3y² -12 z²

= 3 ( x² +2xy+y²) -12z²

= 3 (x+y)²  - 12 z²

= 3 [(x+y)² - 4z² ]

= 3( x + y - 2z )( x + y + 2z)

d) x² - 5 -6 = x² -11

= (x - √11)(x+√11)

a, \(15x^2+10x=5x\left(3x+2\right)\)

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

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

c, \(3x^2+6xy+3y^2-12z^2\)

\(=3\left(x^2+2xy+y^2-4z^2\right)\)

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

d, \(x^2-5x-6=x^2-6x+x-6=x\left(x+1\right)-6\left(x+1\right)=\left(x-6\right)\left(x+1\right)\)

\(x^2+y^2-x^2y^2+xy-x-y\)

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

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

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

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

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

Ngại viết đặt cho nhanh :>

\(a^2-2a+b^2+4b+4c^2-4c+6\)

\(=\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)\)

\(=\left(a-1\right)^2+\left(b+2\right)^2+\left(2c+1\right)^2\)(*)

Ta có : (*) \(\ge0\)

Dấu ''='' xảy ra <=> \(a=1;b=-2;c=-\frac{1}{2}\)(**)

Vậy GTNN biểu thức là 0 <=> ta có (**) 

tìm giá trị nhỏ nhất giúp ạ

Ta có : \(^{4x^2}\)+ \(\frac{1}{x^2}\)+ \(7\)\(=\)\(8x\)+ \(\frac{4}{x}\)

\(\Rightarrow\) \(4x^4\)+ \(1\)+ \(7x^2\)\(=\)\(8x^3\)+ \(4x\)

\(\Rightarrow\) \(4x^4\)- \(8x^3\)+ \(7x^2\)-\(4x\)+\(1\)\(=\)\(0\)

\(\Rightarrow\) 4x³(x-1) - 4x²(x-1) + 3x(x-1) - (x-1) = 0

\(\Rightarrow\) (x-1) ( 4x³ -4x² +3x -1) = 0

\(\Rightarrow\) (x-1) [2x²(2x-1) - x(2x-1) + (2x-1)] = 0

\(\Rightarrow\) (x-1) (2x-1) (2x² - x +1) = 0

\(\Rightarrow\) (x-1) (2x-1) [ 2(x-\(\frac{1}{4}\))² + \(\frac{7}{8}\)] = 0

Dễ thấy : 2(x-\(\frac{1}{4}\))² + \(\frac{7}{8}\)> 0 \(\forall x\)

\(\Rightarrow\) \(\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

Vậy \(x\in\){1;\(\frac{1}{2}\)}