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13 tháng 10 2018

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

\(=x^6-3x^4+9x^2+3x^4-9x^2+27-x^3-6x^2-9x-3x^2-18x-27\)

\(=x^6+\left(-3x^4+3x^4\right)+\left(9x^2-9x^2-6x^2-3x^2\right)+\left(27-27\right)-x^3+\left(-9x-18x\right)\)

\(=x^6-6x^2-3x^2-x^3-27x\)

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

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

24 tháng 9 2021

\(i,=\left(x-3\right)\left(x+3\right)^2-\left(x-3\right)\left(x^2+3x+9\right)\\ =\left(x-3\right)\left(x^2+6x+9-x^2-3x-9\right)\\ =3x\left(x-3\right)=3x^2-9x\\ ii,=x^3-8-25-x^3=-33\)

ii: Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+25\right)\)

\(=x^3-8-x^3-25\)

=-33

31 tháng 12 2020

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

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

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

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

25 tháng 7 2021

a, `(x-3)(x^2+3x+9)-(x^2-1)(9x+27)`

`=x^3-3^3-(9x^3+27x^2-9x-27)`

`=x^3-3^3-9x^3-27x^2+9x+27`

`=-8x^3-27x^2+9x`

b, `(x-2)(x^2+2x+4)-x(x-3)(x+3)`

`=x^3-2^3-x(x^2-9)`

`=x^3-8-x^3+9x`

`=9x-8`

a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(9x+27\right)\)

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

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

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

b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)

\(=x^3-8-x\left(x^2-9\right)\)

\(=x^3-8-x^3+9x\)

\(=9x-8\)

22 tháng 10 2023

1:

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

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)

\(=\left(-3\right)^2+5\left(2x-3\right)\)

\(=9+10x-15=10x-6\)

2: 

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

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

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

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

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)