a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

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

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

\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)

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

\(=\dfrac{x+1}{2x}\)

b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:

\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)

`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`

\(\left(x-2\right)^3+\left(2x+1\right)^2+2\left(x+2\right)\left(1-x\right)-9x^3+2x\)

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

\(=9x^3+6x^2+18x-7+2\left(x-x^2+2-2x\right)-9x^3+2x\)

\(=6x^2+20x-7-2x^2-2x+4=4x^2+18x-3\)

Trả lời:

( x - 2 )³ + ( 2x + 1 )³ + 2 ( x + 2 )( 1 - x ) - 9x³ + 2x

= x³ - 6x² + 12x - 8 + 8x³ + 12x² + 6x + 1 + 2 ( x - x² + 2 - 2x ) - 9x³ + 2x 

= 9x³ + 6x² + 18x - 7 + 2x - 2x² + 4 - 4x - 9x³ + 2x 

= 4x² + 18x - 3 

a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)

\(=-18x^3-46x^2-8x+16\)

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

\(=6x^2+3x+2x^2+2x-3x-3\)

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

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

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

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

\(=\left(x-2\right).1\)

\(=x-2\)

Ta có:

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

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

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

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

\(=\left(x-2\right).1\)

\(=x-2\)

Answer:

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

\(=(4x^2+4x+1)+(4x^2-4x+1)-2(4x^2-1)\)

\(=4x^2+4x+1+4x^2-4x+1-8x^2+2\)

\(=(4x^2+4x^2-8x^2)+(4x-4x)+(1+1+2)\)

\(=4\)

\((x-1)^3-(x+2)(x^2-2x+4)+3(x-1)(x+1)\)

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

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

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

\(=3x-12\)