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

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

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

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

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

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

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

A=\(\frac{-y\left(1-y\right)+x\left(1-y\right)\left(1+y\right)}{\left(1-y\right)}\)

A=\(\frac{\left(1-y\right)\left(-y+x+xy\right)}{1-y}\)=\(x-y+xy\)

\(\frac{\left(2x^2+2x\right)\left(x-3\right)^2}{x\left(x^2-9\right)\left(x+1\right)}.ĐKXD:x\ne3,x\ne0,x\ne-1\)

\(=\frac{2x\left(x+1\right)\left(x-3\right)\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)\left(x+1\right)}=\frac{2x\left(x-3\right)}{x\left(x+3\right)}\)

\(=\frac{2x-6}{x+3}\)

b) Với x=0,5=>\(P=\frac{-5}{3,5}\)

\(P=0\Leftrightarrow2x-6=0\Leftrightarrow x=3\)

\(27+27x+9x^2+x^3\)

\(=x^3+9x^2+27x+27\)

\(=x^3+3x^2+6x^2+18x+9x+27\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+9\right)\)

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

Nếu nhìn kĩ thì bạn sẽ thấy đây là hằng đẳng thức nhé !

\(x^3+9x^2+27x+27=x^3+3.3.x^2+3.3^2.x+3^3\)

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

Đề yêu cầu gì vậy bạn???

\(27+27x+9x^2+x^3\)

\(=x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3\)

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

(Hình bạn tự vẽ nhé) 

a) 

     BE = AC mà AC = DB (2 đường chéo bằng nhau trong hình chữ nhật)

=> BE = BD => Tam giác BDE cân ở B

b) 

CM: \(\Delta ADC=\Delta CBA\left(c.c.c\right)\)

=> S_ADC=S_CBA

mà S_ADC= 1/2. DK.AC

      S_CBA= 1/2. BH.BD

       AC = BD

=> DK = BH mà \(DK\perp BH\)

=> BHDK là hình bình hành

136²  + 36² -72.136

= 136² + 36² - 2.36.136

= (136 - 36)²

= 100² = 10000

\(136^2+36^2-72.136\)

\(=136^2+36^2-2.36.136\)

\(=\left(136-36\right)^2\)

\(=100^2=10000\)

sai đề phải ko

dấu= thành dấu+ạ

\(56^2+44^2+2.44.56\)

\(=\left(56+44\right)^2\)

\(=100^2=10000\)

= (56+44)^2=100^2=10000

hok tốt na