`Answer:`

`a)`

`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`

`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`

`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`

`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`

`=>A=-2x^2+28x-6`

`b)`

`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`

`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`

`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`

`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`

Thay `x=-7` vào ta được:

`B=10(-7)^2-2(-7)^3-7(-7)-6`

`=>B=10.49-2(-343)+49-6`

`=>B=490+686+49-6`

`=>B=1219`

ĐKXĐ: `x>=0;x\ne9`

`(x^2-3)/(sqrtx-3)=((x-sqrt3)(x+sqrt3))/(x+sqrt3)=x-sqrt3`

`a)(x^2-3)/(x+\sqrt3)`

`->` ĐKXĐ : `x>=0;x\ne9`

`=((x-\sqrt3)(x+\sqrt3))/(x+\sqrt3)`

`=(x-\sqrt3)/1`

`=x-\sqrt3`

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

a, ( x+ 2 )( x - 2 ) - ( x-3 ( x-1)

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

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

= 2x -1

a) \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)

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

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

\(=2x-1\)

Bạn Trần Huyền Trang làm sai rồi.

a) \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)

\(=\left(3-xy^2+2+xy^2\right)\left(3-xy^2-2-xy^2\right)\)

\(=5.\left(-2xy^2\right)\)

\(=-10xy^2\)

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

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

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

\(=x^3-3x^2.3+3x.3^2-3^3+2^3-3.2^2.x+3.2.x^2-x^3\)

\(=x^3-9x^2+27x-27+8-12x+6x^2-x^3\)

\(=\left(x^3-x^3\right)+\left(-9x^2+6x^2\right)+\left(27x-12x\right)+\left(-27+8\right)\)

\(=-3x^2+15x-19\)

a) Ta có: \(\left(x-2\right)^3-\left(3+x^2\right)\left(3-x\right)\)

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

\(=x^3-6x^2+12x-8+x^3+3x-3x^2-9\)

\(=2x^3-9x^2+15x-17\)

b) Ta có: \(x\left(x-14\right)-10\left(x-1\right)^2\)

\(=x^2-14x-10\left(x^2-2x+1\right)\)

\(=x^2-14x-10x^2+20x-10\)

\(=-9x^2+6x-10\)

c) Ta có: \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)\)

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

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

\(=x^2+4x+4\)

d) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3-27\right)\)

\(=x^3-27-x^3+27\)

=0

Trả lời:

( x - 3 ) ( x² + 3x + 9 ) - ( x² - 27x )

= x³ - 27 - x² + 27x

= x³ - x² + 27x - 27

  ( 3 ----- 3)-----no biết       
 

A = (2x - 1)(x + 2) - 3x² + (x - 1)²

= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1

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

= x - 1

B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)

= x³ - 8 - x³ - x² - 9 + x²

= (x³ - x³) + (-x² + x²) + (-8 - 9)

= -17

A = (2x - 1)(x + 2) - 3x² + (x - 1)²

= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1

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

= x - 1

B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)

= x³ - 8 - x³ - x² - 9 + x²

= (x³ - x³) + (-x² + x²) + (-8 - 9)

= -17

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