Đặt x + 2 = t ta có  biểu thứ mới : 

\(t\left(t+1\right)^2\left(t+2\right)-12=t\left(t^2+2t+1\right)\left(t+2\right)-12\)

\(=\left(t^3+2t^2+t\right)\left(t+2\right)-12=t^4+2t^3+2t^3+4t^2+t^2+2t-12\)

\(=t^4+4t^3+5t^2+2t-12=\left(t-1\right)\left(t^3+5t^2+10t+12\right)\)

\(=\left(t-1\right)\left(t+3\right)\left(t^2+2t+4\right)=\left(x+1\right)\left(x+5\right)\left[\left(x+2\right)^2+2\left(x+2\right)+4\right]\)

( x + 2 )( x + 3 )²( x + 4 ) - 12

= [ ( x + 2 )( x + 4 ) ]( x + 3 )² - 12

= ( x² + 6x + 8 )( x² + 6x + 9 ) - 12

Đặt t = x² + 6x + 8

= t( t + 1 ) - 12

= t² + t - 12

= t² - 3t + 4t - 12

= t( t - 3 ) + 4( t - 3 )

= ( t - 3 )( t + 4 )

= ( x² + 6x + 8 - 3 )( x² + 6x + 8 + 4 )

= ( x² + 6x + 5 )( x² + 6x + 12 )

= ( x² + 5x + x + 5 )( x² + 6x + 12 )

= [ x( x + 5 ) + ( x + 5 ) ]( x² + 6x + 12 )

= ( x + 5 )( x + 1 )( x² + 6x + 12 )

\(a^7-b^7=\left(a-b\right)\left(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6\right)\)

\(a^7-b^7=\left(a-b\right)\left(a^6+a^5b+a^4b^2+a^3b^c+a^2b^4+ab^5+b^6\right)\)

\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow x+y+z=0\).

\(P=\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz}{\left(-x\right)\left(-y\right)\left(-z\right)}=-1\)

pt bậc hai à?

\(x=2\)

XL mik đag bận nên không giải chi tiết cho bn đc!!!

\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right)\div\left(\frac{1}{x}+x-2\right)\)

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)

\(=\left(\frac{1}{x\left(x+1\right)}-\frac{2-x}{x+1}\right)\div\left(\frac{1}{x}+\frac{x^2}{x}-\frac{2x}{x}\right)\)

\(=\left(\frac{1}{x\left(x+1\right)}-\frac{x\left(2-x\right)}{x\left(x+1\right)}\right)\div\left(\frac{x^2-2x+1}{x}\right)\)

\(=\left(\frac{1}{x\left(x+1\right)}-\frac{2x-x^2}{x\left(x+1\right)}\right)\times\frac{x}{\left(x-1\right)^2}\)

\(=\left(\frac{1-2x+x^2}{x\left(x+1\right)}\right)\times\frac{x}{\left(x-1\right)^2}\)

\(=\frac{\left(x-1\right)^2}{x\left(x+1\right)}\times\frac{x}{\left(x-1\right)^2}\)

\(=\frac{1}{x+1}\)

 SOORY TUI HỌC LỚP  7 NHA

a) \(\frac{2x-2y}{x^2-y^2}=\frac{2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{2}{x+y}\)

\(\frac{5}{2x^2+4xy+2y^2}=\frac{5}{2\left(x^2+2xy+y^2\right)}=\frac{5}{2\left(x+y\right)^2}\)

MTC : 2( x + y )²

=> \(\hept{\begin{cases}\frac{2x-2y}{x^2-y^2}=\frac{2}{x+y}=\frac{2\times2\left(x+y\right)}{\left(x+y\right)\times2\left(x+y\right)}=\frac{4x+4y}{2\left(x+y\right)^2}\\\frac{5}{2x^2+4xy+2y^2}=\frac{5}{2\left(x^2+2xy+y^2\right)}=\frac{5}{2\left(x+y\right)^2}\end{cases}}\)

b) \(\frac{x-y}{x^3-y^3}=\frac{x-y}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\frac{1}{x^2+xy+y^2}\)

\(\frac{5}{2x^2+2x+2}=\frac{5}{2\left(x^2+x+1\right)}\)

\(\frac{6}{4x^3+4x+4}=\frac{6}{4\left(x^2+x+1\right)}=\frac{3}{2\left(x^2+x+1\right)}\)

MTC : 2( x² + x + 1 )( x² + xy + y² )

=> \(\frac{1}{x^2+xy+y^2}=\frac{2\left(x^2+x+1\right)}{2\left(x^2+x+1\right)\left(x^2+xy+y^2\right)}=\frac{2x^2+2x+2}{2\left(x^2+x+1\right)\left(x^2+xy+y^2\right)}\)

=> \(\frac{5}{2\left(x^2+x+1\right)}=\frac{5\left(x^2+xy+y^2\right)}{2\left(x^2+x+1\right)\left(x^2+xy+y^2\right)}=\frac{5x^2+5xy+5y^2}{2\left(x^2+x+1\right)\left(x^2+xy+y^2\right)}\)

=> \(\frac{3}{2\left(x^2+x+1\right)}=\frac{3\left(x^2+xy+y^2\right)}{2\left(x^2+x+1\right)\left(x^2+xy+y^2\right)}=\frac{3x^2+3xy+3y^2}{2\left(x^2+x+1\right)\left(x^2+xy+y^2\right)}\)

a, \(\frac{2x-2y}{x^2-y^2};\frac{5}{2x^2+4xy+2y^2}\)

Ta có :  \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)

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

MTC : \(2\left(x-y\right)\left(x+y\right)^2\)

\(\frac{2x-2y}{x^2-y^2}=\frac{2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{2\left(x-y\right)\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)^2}\)

\(\frac{5}{2\left(x^2+2xy+y^2\right)}=\frac{5}{2\left(x+y\right)^2}=\frac{5\left(x-y\right)}{2\left(x-y\right)\left(x+y\right)^2}\)