đk x khác 2 

\(P=\dfrac{x^2+2x+4-x^2-8-4x+8}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{-2}{x^2+2x+4}\)

\(ĐKXĐ:x\ne2\)

\(P=\dfrac{1}{x-2}-\dfrac{x^2+8}{x^3-8}-\dfrac{4}{x^2+2x+4}\)

\(=\dfrac{x^2+2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{x^2+8}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{4\left(x-2\right)}{x^2+2x+4}\)

\(=\dfrac{x^2+2x+4-x^2-8-4x+8}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{4-2x}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{-2\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=-\dfrac{2}{x^2+2x+4}\)

(2+x).(2-x) = (4-x²)

* hằng đẵng thức đáng nhớ

ý mk là nhân hai cái với nhau  chứ ai cũng bt là = (4-x^2) rồi.

giống kiểu (2+x).(2+x)=4+2x+2x+x^2

a) \(x^2+10x+26+y^2+2y\)

\(=x^2+10x+25+y^2+2y+1\)

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

b) Xem lại đề 

c) \(x^2-2xy+2y^2+2y+1\)

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

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

ra 900

   \(\dfrac{3}{5}\)a³b x (10ab³ - \(\dfrac{5}{3}\)b² + \(\dfrac{5}{6}\)ab)

=\(\dfrac{3}{5}\)a³b . 10ab³ - \(\dfrac{3}{5}\)a³b. \(\dfrac{5}{3}\)b² + \(\dfrac{3}{5}\)a³b . \(\dfrac{5}{6}\)ab

= 6a⁴b⁴ - a³b³ + \(\dfrac{1}{2}\) a⁴b²

Mn giúp với

thk mn