a)

\(2x\left(9-x\right)+\left(2x+5\right)\left(x+1\right)\\ =18x-2x^2+2x^2+5x+2x+5\\ =25x+5\\ =5\left(5x+1\right)\)

b)

\(\left(x-3\right)^2+\left(x+3\right)^2+2\left(3-x\right)\left(3+x\right)\\ =\left(3-x\right)^2+2\left(3-x\right)\left(3+x\right)+\left(3+x\right)^2\\ =\left[\left(3-x\right)+\left(3+x\right)\right]^2\\ =6^2=36\)

`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`

a) (x+3)(x^2-3x+9)-(54+x^3)

= x^3- 3x^2+9x+3x^2-9x+27-54-x63

= -27

b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)

= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]

= [(2x)3^3+ y^3] – [(2x)^3 – y^3]

= (2x)^3 + y^3 – (2x)^3 + y^3

= 2y^3

a)(x+3)(X^2-3x+9)-(54+x^3)

= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)

= 27- 54

= -27

b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)

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

= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)

= \(2y^3\)

a) Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3+27-54-x^3\)

=-27

\(A=\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\left(x>3\right)\\ A=\dfrac{\left(x+3\right)+2\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+3\right)}}\\ A=\dfrac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2+\sqrt{x+3}\right)}\)

Tới đây chịu rùi, hình như đề sai đk?

\(=8x^3+27-8x^3-7=20\)

bằng 20

a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)

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

(\(3+\dfrac{x}{3-x}+\dfrac{2x}{3+x}-\dfrac{4x^2-3x-9}{x^2-9}\) ):\(\left(\dfrac{2}{3-x}-\dfrac{x-1}{3x-x^2}\right)\)\(=\left(\dfrac{3x^2-27}{\left(x-3\right)\left(x+3\right)}+\dfrac{-x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2-3x-9}{\left(x-3\right)\left(x+3\right)}\right)\)\(:\left(\dfrac{2x}{x\left(3-x\right)}-\dfrac{x-1}{x\left(3-x\right)}\right)\)

\(=\dfrac{3x^2-27-x^2-3x+2x^2-6x-4x^2+3x+9}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{-6x-18}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) \(=\dfrac{-6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

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

Bạn gõ cái này nhé 

a: \(A=\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{2x^2-2x-50}{2x\left(x-5\right)}\)

\(=\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\)

\(=\dfrac{x^2-10x+25}{2x\left(x-5\right)}=\dfrac{x-5}{2x}\)

b: Để A=1/3 thì x-5/2x=1/3

=>3x-15=2x

=>x=15

\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)

\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)