Bài 4: Những hằng đẳng thức đáng nhớ (Tiếp)

\(2x^4-2\sqrt{6}x^2y^2+3y^2\)

\(=\left(\sqrt{2x^2}\right)^2-2.\sqrt{2x^2}.\sqrt{3y^2}+\left(\sqrt{3y^2}\right)^2\)

\(=\left(\sqrt{2x^2}-\sqrt{3y^2}\right)^2\)

1: \(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-4x^2+12x-9\)

\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)

\(=-2x^3-13x^2-x-12\)

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

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

\(=-18x^3-48x^2-8x+16\)

\(=64x^3-48x^2+12x-1-64x^3+48x^2-12x+9\\ =-1+9\\ =8\)

\(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)=\left(64x^3-48x^2+12x-1\right)-\left(64x^3+12x-48x^2-9\right)=64x^3-48x^2+12x-1-64x^3-12x+48x^2+9=8\)

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

`x^3 - 3x^2 + 3x - 1 - x^3 - 3x^2 - 3x - 1 + 6x^2 - 6`.

`= -1-1-6=-8`

\(a,x^3+12x^2+48x+64=\left(x+4\right)^3\\ \)

Thay \(x=6\)

\(\left(6+4\right)^3=10^3=1000\)

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

Thay \(x=22\)

\(\left(22-2\right)^3=20^3=8000\)

`x^3 + 12x^2 +48x +64`

` = x^2 + 3*x^2*4 + 3*x*4^2 +4^3`

`= (x+4)^3`

Thay `x=6` vào biểu thức ta đc

`(4+6)^3 = 10^3 = 1000`

`x^3 -6x^2 +12x -8 `

`= x^3 -3*x^2*2 + 3*x*2^2 -2^3`

`= (x-2)^3`

Thay `x=22` vào biểu thức ta đc

`(22-2)^3 = 20^3 = 8000`

Thay  vào biểu thức ta đc

Thay  vào biểu thức ta đc

`a)-x^3+3x^2-3x+1`

`=-(x^3-3x^2+3x-1)`

`=-(x-1)^3`

`=(1-x)^3`

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`b)8-12x+6x^2-x^3`

`=2^3-3.2^2x+3.2x^2-x^3`

`=(2-x)^3`

Đề bài là gì?

a)

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

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

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

b)

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

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

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

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

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

1:

=>(2x-6)^3=(ax+b)^3

=>a=2; b=-6

3:

N=(x-y)^3+3xy(x-y)-3xy

=1+3xy-3xy

=1