\(x^2-y^2-x-y\)

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

\(=\left(x+y\right)\left(x-y-1\right)\)

Dùng hằng đằng thức là ra mà :))

Với |x| = 2 khi x = 2 hoặc x = -2 

a, \(A=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)

\(=6x^2+7x-5+12x^2+5x-2=18x^2+12x-7\)

Với x = 2 thì \(A=89\)

Với x = -2 thì \(A=41\)

b, \(B=\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\)

\(=x^4+x^3y-x^3y-x^2y^2+x^2y^2+xy^3-xy^3-y^4=x^4-y^4\)

Thay x = 2 ; y = -1/4 ta được : \(=16-\frac{1}{256}=\frac{4095}{256}\)

Bài 4 : 

a, \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)

\(\Leftrightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-12x^2+15x-3+12x^2+28x-24=-27\)

\(\Leftrightarrow43x=0\Leftrightarrow x=0\)

b, \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\)

\(\Leftrightarrow-12x-2=-17x+20\Leftrightarrow5x=22\Leftrightarrow x=\frac{22}{5}\)

( x + 3 )( x + 4 )( x + 5 )( x + 6 ) + 1 

= [ ( x + 3 )( x + 6 ) ][ ( x + 4 )( x + 5 ) ] + 1

= ( x² + 9x + 18 )( x² + 9x + 20 ) + 1

= ( x² + 9x + 19 - 1 )( x² + 9x + 19 + 1 ) 1 

= ( x² + 9x + 19 )² - 1² + 1 = ( x² + 9x + 19 )²

x² - 2x( y + 2 ) + y² + 4y + 4

= x² - 2x( y + 2 ) + ( y + 2 )²

= ( x - y - 2 )²

1/A = x² + 3x - 2 = ( x² + 3x + 9/4 ) - 17/4 = ( x + 3/2 )² - 17/4 ≥ -17/4 ∀ x

Dấu "=" xảy ra <=> x = -3/2 => MinA = -17/4

2/ a) = 7x( y + z )

b) = ( x - 2 )² - (5y)² = ( x - 5y - 2 )( x + 5y - 2 )

3/ a) <=> ( x - 5 )( x + 5 ) - ( x + 5 ) = 0

<=> ( x + 5 )( x - 6 ) = 0 <=> x = -5 hoặc x = 6

b) <=> x² - 3x - x + 3 = 0 <=> ( x - 1 )( x - 3 ) = 0 

<=> x = 1 hoặc x = 3

c) x - ( x + 1 )² = 0 <=> x - x² - 2x - 1 = 0

<=> x² + x + 1 = 0 dễ thấy pt trên vô nghiệm

Điều kiện: \(x\ne1,x\ne0\).

\(E=\left[\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{1-x}\right]\div\frac{2x}{x^3+x}\)

\(=\left(\frac{x^2-2x+1}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right)\div\frac{2}{x^2+1}\)

\(=\left[\frac{\left(x-1\right)^3-\left(1-2x^2+4x\right)+\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right].\frac{x^2+1}{2}\)

\(=\frac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{2}\)

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

\(4E=2\left(x^2+1\right)=x+8\)

\(\Leftrightarrow2x^2-x-6=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)(thỏa mãn) 

\(E=\frac{x^2+1}{2}>0\)do \(x^2\ge0\).

\(\left(2x-1\right)^3-8\left(x-3\right)\left(x+3\right)+12x\left(x-2\right)\)

\(=8x^3-12x^2+6x-1-8\left(x^2-9\right)+12x^2-24x\)

\(=8x^3-18x-1-8x^2+72=8x^3-8x^2-18x+71\)