mn giúp e với em quên cách làm rồi

BCNN(28; 30; 2)

28 = 2\(^2\).7; 30 = 2.3.5; 2 = 2

BCNN(28; 30; 2) = 2\(^2.3.5.7\) = 420

BCNN(25; 30; 3)

25 = 5\(^2\) ; 30 = 2.3.5; 3 = 3

BCNN(25; 30; 3) = 2.5\(^2\).3 = 150

BCNN(30; 35; 6; 3)

30 = 2.3.5; 35 = 5.7; 6 = 2.3; 3 = 3

BCNN(30; 35; 6; 3) = 2.3.5.7 = 210

   4x² - y² - 12x + 9

= [4x² - 12x + 9]- y²

= [(2x)² - 2.2.x.3 + 3²] - y²

= (2x - 3)² - y²

= (2x - 3 - y).(2x  - 3 + y)

a; 2x - 8  = 2(x  -4)

b; 4 - 6x = 2(2 - 3x)

c; 4xy - 2y² = 2y(2x - y)

d; 3x² - 12  = 3(x² - 4) = 3(x - 2)(x + 2)

e; 36x² - 9y² 

= 9(4x² - y²)

= 9.(2x - y).(2x + y)ư

f; 6x( x - 1) + 3.(x - 1)

 = 3.(x - 1).(6x + 1)

g; x² - 4x + 4 - 4y²

= (x² - 4x + 4 ) - 4y²

= (x - 2)² - 4y²

= (x - 2 - 2y)(x - 2 + 2y)

\(x^2+x-1=0\)

=>\(x^2+x+\dfrac{1}{4}-\dfrac{5}{4}=0\)

=>\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

=>\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}}{2}-\dfrac{1}{2}\\x=-\dfrac{\sqrt{5}}{2}-\dfrac{1}{2}\end{matrix}\right.\)

   \(x^3\) + 3\(x^2\)y + 3\(xy^2\) + y³ - \(x-y\)

= (\(x^3\) + 3\(x^2\)y + 3\(xy^2\) + y³) - (\(x+y\))

= (\(x+y\))³ - (\(x+y\))

= (\(x+y\))[(\(x+y\))² - 1]

= (\(x+y\))[\(x+y-1\)][\(x+y+1\)]

cứu tuii ii mn :<

`3 - (x + 1)^2`

`= 3 - (x^2 + 2x + 1)`

`= 3 - x^2 - 2x - 1`

`= 2 - x^2 - 2x`

  3 - (\(x+1\))^\(2\)

= 3 - (\(x^2\) + 2\(x\) + 1)

= 3 - \(x^2\) - 2\(x-1\)

= - \(x^2\) - 2\(x\) + (3 -1)

= - \(x^2\) - 2\(x\) + 2 

\(x^2\) x \(x\) = \(x^{2+1}\) = \(x^3\) 

cảm ơn ạ 

\(x^2\) - \(x\) + 3.(\(x-1\)) = 0

\(x\left(x-1\right)\) + 3(\(x-1\)) = 0

(\(x-1\))\(\left(x+3\right)\) = 0

\(\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; 1}

C = 1 - \(\dfrac{\left(x+3\right)\left(x+5\right)}{2}\)

C = 1 - \(\dfrac{x^2+5x+3x+15}{2}\)

C = 1 - \(\dfrac{x^2+\left(5x+3x\right)+15}{2}\)

C = 1 - \(\dfrac{x^2+8x+16-1}{2}\)

C = 1 - \(\dfrac{\left(x^2+2.x.4+4^2\right)}{2}\) + \(\dfrac{1}{2}\)

C = (1 + \(\dfrac{1}{2}\)) -  \(\dfrac{\left(x+4\right)^2}{2}\)

C =  \(\dfrac{3}{2}\)- \(\dfrac{\left(x+4\right)^2}{2}\)

Vì (\(x+4\))² ≥ 0 \(\forall\) \(x\) ⇒ - \(\dfrac{1}{2}\)(\(x+4\))² ≤ 0 ∀ \(x\)

    ⇒ \(\dfrac{3}{2}\) - \(\dfrac{\left(x+4\right)^2}{2}\) ≤ \(\dfrac{3}{2}\) dấu bằng xảy ra khi \(x+4\) = 0 ⇒ \(x=-4\)

Vậy giá trị lớn nhất của biểu thức C là \(\dfrac{3}{2}\) xảy ra khi \(x=-4\) 

    \(x^3\)  - \(x-y\) + y³

= (\(x^3\) + y³) - (\(x+y\))

= (\(x+y\)).(\(x^2\) - \(xy\) + y²) - (\(x+y\))

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

\(x^3-x-y+y^3\)

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

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

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