\(\left(6-x\right)^2=x-6\)\(< =>\left(6-x\right)^2+6-x=0\)

\(< =>\left(6-x\right)\left(6-x+1\right)=0\)

\(< =>\orbr{\begin{cases}x=6\\x=7\end{cases}}\)

Trả lời:

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

\(\Leftrightarrow\left(x-6\right)-\left(6-x\right)^2=0\)

\(\Leftrightarrow\left(x-6\right)-\left(x-6\right)^2=0\)

\(\Leftrightarrow\left(x-6\right)\left(1-x+6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(7-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}}\)

Vậy x = 6; x = 7 là nghiệm của pt.

Trả lời:

Ta có: \(A=\left(x^2+x\right)^2+5y\left(x^2+x\right)+6y^2\)

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

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

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

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

Trả lời:

+) \(x^4+3x^3-9x-9\)

\(=\left(x^4-9\right)+\left(3x^3-9x\right)\)

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

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

+) \(x^4+3x^3-9x-27\)

\(=\left(x^4+3x^3\right)-\left(9x+27\right)\)

\(=x^3\left(x+3\right)-9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^3-9\right)\)

Bài 1. 

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

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

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

d) \(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

Trả lời:

a, \(2x^2-2xy-3x+3y\)

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

\(=2x\left(x-y\right)-3\left(x-y\right)\)

\(=\left(x-y\right)\left(2x-3\right)\)

b, \(-x^2-y^2+2xy+16\)

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

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

\(=\left(4-x+y\right)\left(4+x-y\right)\)

c, \(y^2-x^2+2yz+z^2\)

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

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

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

d, \(3x^2-6xy+3y^2-12z^2\)

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

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

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

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

ta có :

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

Vậy \(x^4-1-\left(2x^3-2x\right):\left(x^2-1\right)=x^2-2x+1\)

x² + 6x - y² + 9 = 

= ( x² + 6x + 9 ) - y² 

= ( x + 3 )² - y² 

= ( x + 3 - y ) . ( x + 3 + y )

Hok tốt!!!!!!!!!!!

ta có 

a.\(N=\left(10x\right)^2-2.\left(10x\right)+1=\left(10x-1\right)^2=\left(3-1\right)^2=4\text{ khi }x=\frac{3}{10}\)

b.\(P=\left(5x\right)^2-2.5xy^2+y^4=\left(5x-y^2\right)^2=\left(5.5-4^2\right)=9^2=81\)

1+1=3-1=2 nha bạn

\(A=\left(x^2-5x-4\right)^2+7\left(x^2-5x-4\right)+10\)

Đặt \(t=x^2-5x-4\).

\(A=t^2+7t+10=t^2+2t+5t+10=\left(t+2\right)\left(t+5\right)\)

\(=\left(x^2-5x-4+2\right)\left(x^2-5x-4+5\right)\)

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

\(x^2-16+2\left(4-x\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)-2\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4-2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)

Trường hợp 1: \(x-4=0\Leftrightarrow x=4\)

Trường hợp 2: \(x+2=0\Leftrightarrow x=-2\)

Vậy: \(S=\left\{4;-2\right\}\)