A=1002
B=5

\(x^3+3x^2+6x+4=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\)

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

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

a)   \(x^3+3x^2+6x+4\)

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

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

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

b) \(3a^2c^2+bd+3abc+acd\)

\(=\left(3a^2c^2+acd\right)+\left(3abc+bd\right)\)

\(=ac\left(3ac+d\right)+b\left(3ac+d\right)\)

\(=\left(ac+b\right)\left(d+3ac\right)\)

c)  \(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)\)

\(=3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)

d) \(x^2+y^2-x^2y^2+xy-x-y\)

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

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

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

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

e)  \(a^6-b^6=\left(a^3-b^3\right)\left(a^3+b^3\right)=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)

a ) \(x^3+3x^2+6x+4\)

\(=x^3+3x^2+3x+1+3x+3\)

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

\(=\left(x+1\right)\left[\left(x+1\right)^2+3\right]\)

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

b ) \(3a^2c^2+bd+3abc+acd\)

\(=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)\)

\(=3ac\left(ac+b\right)+d\left(ac+b\right)\)

\(=\left(3ac+d\right)\left(ac+b\right)\)

c ) \(3a^2-6ab+3b^2-12c^2\)

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

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

\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)

\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)

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

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

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

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

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

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

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

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

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

e ) \(a^6-b^6=\left(a^3\right)^2-\left(b^3\right)^2=\left(a^3-b^3\right)\left(a^3+b^3\right)\) \(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)

a, \(x^3+3x^2+6x+4\)

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

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

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

e, \(a^6-b^6\)

\(=\left(a^3\right)^2-\left(b^3\right)^2\)

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

a) x³ - 4x² + 4x

= x(x² - 4x + 4)

= x(x - 2)²

b) 2xy - x² - y² + 16

= 16 -x² + 2xy - y²

= 16 - (x² - 2xy + y²)

= 4² - (x - y)²

= [4 - (x - y)].(4 + x - y)

= (4 - x + y)(4 + x - y)

c) x² - y² - 2yz - z²

= x² - (y² + 2yz + z²)

= x² - (y + z)²

= [x -(y + z)].(x + y +z)

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

d) 3a² - 6ab + 3b² - 12c²

= 3(a² - 2ab + b² - 4c²)

= 3[(a2 - 2ab + b²) - (2c)²]

= 3[(a - b)² - (2c)²]

= 3(a - b - c)(a - b + c)

con D bạn chép sai đề bài rồi, phải là +3b2 chứ. tích cho mik nha, ko thì lần sau mik ko giúp đâu ihihihi.....!!!!!!!!!

\(\dfrac{27x^3}{y^3}+\dfrac{8y^3}{125}\left(y\ne0\right)\\ =\left(\dfrac{3x}{y}\right)^3+\left(\dfrac{2y}{5}\right)^3\\ =\left(\dfrac{3x}{y}+\dfrac{2y}{5}\right)\left(\dfrac{9x^2}{y^2}-\dfrac{6x}{5}+\dfrac{4y^2}{25}\right)\)

Ta có: \(\dfrac{27x^3}{y^3}+\dfrac{8y^3}{125}\)

\(=\left(\dfrac{3x}{y}\right)^3+\left(\dfrac{2y}{5}\right)^3\)

\(=\left(\dfrac{3x}{y}+\dfrac{2y}{5}\right)\cdot\left(\dfrac{9x^2}{y^2}-\dfrac{6xy}{5y}+\dfrac{4y^2}{25}\right)\)

\(=\left(\dfrac{3x}{y}+\dfrac{2y}{5}\right)\left(\dfrac{9x^2}{y^2}-\dfrac{6x}{5}+\dfrac{4y^2}{25}\right)\)

a) \(ĐKXĐ:2x^2+6x+1\ge0\)

Với \(x\ge2\) pt cho trở thành :

\(2x^2+6x+1=x^2+4x+4\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=3\) ( do \(x\ge2\) )

Vậy pt có tập nghiệm \(S=\left\{3\right\}\)

\(a.\sqrt{2x^2+6x+1}=x+2\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\2x^2+6x+1=x^2+4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x=1\\ \Rightarrow S=\left\{1\right\}\)

\(b.\) ĐKXĐ: \(y\ne0\)\(\left(I\right)\Rightarrow x+\dfrac{1}{y}=\dfrac{x}{y}+\dfrac{1}{y}\Leftrightarrow x=\dfrac{x}{y}\Leftrightarrow x\left(1-\dfrac{1}{y}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\y=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{1}{2}\\x=1\end{matrix}\right.\left(TM\right)\Rightarrow S=\left\{\left(0;\dfrac{1}{2}\right);\left(1;1\right)\right\}\)