\(\begin{array}{l}(x - 4) + \left[ {({x^2} + 2x) + (7 - x)} \right]\\ = x - 4 + ({x^2} + 2x + 7 - x)\\ = x - 4 + {x^2} + 2x + 7 - x\\ = {x^2} + (x + 2x - x) + ( - 4 + 7)\\ = {x^2} + 2x + 3\end{array}\)

a: =9x^2-12x+4-4x^2+14x

=5x^2+2x+4

b: \(=\dfrac{2+x+1+x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x+2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

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

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(a.x\left(2x+5\right)=2x^2+5x\)

\(b.3x^2\left(5x^3-2x+9\right)=15x^5-6x^3+27x^2\)

\(c.\left(x+2\right)\left(4x^3+x-7\right)=4x^4+x^2-7x+8x^3+2x-14=4x^4+8x^3+x^2-5x-14\)

\(d.\left(16x^2+12x^3y^4-4x^2y\right):4x^2y=4x^3+3x^2y^3-1\)

a: \(=\dfrac{5}{2x^2y}+\dfrac{2}{3xy}-\dfrac{y}{x^3}\)

\(=\dfrac{5\cdot3\cdot x}{6x^3y}+\dfrac{2\cdot2\cdot x^2}{6x^3y}-\dfrac{6y^2}{6x^3y}\)

\(=\dfrac{15x+4x^2-6y^2}{6x^3y}\)

b: \(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)

c: \(=\dfrac{x^4-1-x^4+3x^2}{x^2-1}=\dfrac{3x^2-1}{x^2-1}\)

A=x^4+3x^3-5x^2+7

B=x^2+4x^2+2x+1=5x^2+2x+1

A-B=x^4+3x^3-5x^2+7-5x^2-2x-1

=x^4+3x^3-10x^2-2x+6

\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)

\(=x^3-8-\left(x^3-3x^2+3x-1\right)+7\)

\(=x^3-8+7-x^3+3x^2-3x+1\)

\(=\left(x^3-x^3\right)+\left(7+1-8\right)+3x^2-3x\)

\(=3x^2-3x=3x\left(x-1\right)\)

\(x\left(x+2\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

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

\(=x\left(4-x^2\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=4x-x^3+\left(x^3+9\right)\)

\(=4x-\left(x^3-x^3\right)+9\)

\(=4x+9\)

( x - 2 )( x² + 2x + 4 ) - ( x - 1 )³ + 7

= x³ - 8 - ( x³ - 3x² + 3x - 1 ) + 7 

= x³ - 8 - x³ + 3x² - 3x + 1 + 7

= 3x² - 3x = 3x( x - 1 )

x( x + 2 )( 2 - x ) + ( x + 3 )( x² - 3x + 9 )

= ( x² + 2x )( 2 - x ) + x³ + 27

= 2x² - x³ + 4x - 2x² + x³ + 27

= 4x + 27