\(\left|x-2009\right|=x-2009\) 

\(\Rightarrow x-2009=2009-x\Leftrightarrow x=2009\)

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

Vậy ta có đpcm 

\(9x^2-2.3x.5+25+16y^2+8y+1=0\)

\(\Leftrightarrow\left(3x-5\right)^2+\left(8y+1\right)^2=0\)

mà \(\left(3x-5\right)^2+\left(8y+1\right)^2\ge0\)

Dấu ''='' xảy ra khi x = 5/3 ; y = -1/8 

\(a^3+2a^2-a-2\) 

= \(a^2\left(a+2\right)-\left(a+2\right)\)

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

=\(\left(a+2\right)\left(a-1\right)\left(a+1\right)\)
   

a^3 + 2a^2 -a -2
= a^2. (a+2) + (a+2)
= (a+2) . (a^2 - 1)
= (a+2) . (a+1) . (a-1)

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