\(2x\left(x^2-3y\right)=2x^3-6xy\)

\(-18x^3y^4:3xy^4=-\left(6x^2\right)\)

a) \(2x\left(x^2-3y\right)\)

\(=2x^3-6xy\)

ta có: H = 4x - x^2 = - (x^2 -4x) = -(x^2-4x+4-4) = -(x-2)^2 + 4

mà \(-\left(x-2\right)^2+4\le4\)

Để H có GTLN

=> -(x-2)^2 + 4 = 4

-(x-2)^2 = 0

=> x - 2 = 0 => x = 2

KL:...

Ta có: \(H=4x-x^2\)

\(\Rightarrow H=-x^2+4x\)

\(\Rightarrow H=-x^2+4x-4+4\)

\(\Rightarrow H=-\left(x-2\right)^2+4\)

Ta thấy: \(-\left(x-2\right)^2\le0\)với mọi x

\(\Leftrightarrow-\left(x-2\right)^2+4\le4\)với mọi x

Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\)

                        \(\Leftrightarrow x=2\)

Vậy................

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???

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

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

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