Ta có \(f\left(x\right)=\left(x+1\right)\left(x-3\right)\left(x+5\right)\left(x+9\right)+256\)

\(f\left(x\right)=\left(x+1\right)\left(x+5\right)\left(x-3\right)\left(x+9\right)+256\)

\(f\left(x\right)=\left(x^2+6x+5\right)\left(x^2+6x-27\right)+256\)

\(f\left(x\right)=\left[\left(x^2+6x-11\right)^2-256\right]+256\)

\(f\left(x\right)=\left(x^2+6x-11\right)^2\)

4.(\(x\) + 1)² + (2\(x\) - 1)² - 8.(\(x\) - 1)(\(x-1\) )= 11

4\(x^2\) + 8\(x\) + 4 + 4\(x^2\) - 4\(x\)  + 1 - 8\(x^2\) + 16\(x\)  - 8 = 11

(4\(x^2\) + 4\(x^2\) - 8\(x^2\)) + (8\(x\) - 4\(x\) + 16\(x\)) + (4 + 1 - 8) = 11

                               20\(x\)  - 3 = 11

                               20\(x\)        = 11 + 3

                               20\(x\)        = 14

                                   \(x\)        = \(\dfrac{7}{10}\)

Lời giải:

$(x+1)^3-(x-1)^3-6(x-1)^2=-10$
$\Leftrightarrow (x^3+3x^2+3x+1)-(x^3-3x^2+3x-1)-6(x^2-2x+1)=-10$

$\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10$

$\Leftrightarrow (x^3-x^3)+(3x^2+3x^2-6x^2)+(3x-3x+12x)+(1+1-6)=-10$

$\Leftrightarrow 12x-4=-10$

$\Leftrightarrow 12x=-10+4=-6$

$\Leftrightarrow x=\frac{-6}{12}=\frac{-1}{2}$

\(-x^2+6x+6y+y^2\\=-(x^2-6x+9)+(y^2+6y+9)\\=-(x-3)^2+(y+3)^2\\=(y+3)^2-(x-3)^2\\=[(y+3)-(x-3)][(y+3)+(x-3)]\\=(y+3-x+3)(y+3+x-3)\\=(y-x+6)(x+y)\)

\(x^2-3x+xy-3y\\=(x^2-3x)+(xy-3y)\\=x(x-3)+y(x-3)\\=(x-3)(x+y)\)

\(225-4x^2-4xy-y^2\\=225-(4x^2+4xy+y^2)\\=225-[(2x)^2-2\cdot2x\cdot y+y^2]\\=15^2-(2x-y)^2\\=[15-(2x-y)][15+(2x-y)]\\=(15-2x+y)(15+2x-y)\)

sos

sos

=x mũ 3 + 8 -( 3x mũ 2 + 6x)

= (x+ 2) (x mũ 2 - 2x + 4) -3x ( x+2)

= (x + 2) (x mũ 2 -2x + 4 - 3x)

(x+2)(x mũ 2 -5x + 4)