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

\(x^2-2xy+y^2+15\\ =\left(x-y\right)^2+15\ge15\forall x,y\)

Dấu bằng xảy ra khi

\(\left(x-y\right)^2=0\\ =>x-y=0\\ =>x=y\)

a) x³ + 64 = x³ + 4³ = x³ + 4³ = (x + 4) (x² - 64x + 16) 

b) (x - 2)³ + (x + 1)³ = x³ - 3.x².2 + 3.x.2² - 2³ + x³ + 3.x².1 + 3.x.1² + 1³

= x³ - 6x² + 12x - 8 + x³ + 3x² + 3x + 1

= x⁶ - 3x⁴ + 15x² - 7

c)(3x + 4) (9x² - 12x + 16) = (3x + 4) [(3x)² - 3x.4 + 4²] = 3x³ + 4²

mọi ng giúp với

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

\(\Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\)

\(\Leftrightarrow5x+6=1\)

\(\Leftrightarrow5x=-5\)

\(\Leftrightarrow x=-1\)

(3x - 1)(3x + 1) - (x - 5)(9x - 1) - 25x

= 9x² - 1 - 9x² + x + 45x - 5 - 25x

= 21x - 6

Bậc của đa thức sau khi thu gọn là 1

`= 9x^2 - 1 - 9x^2 + 45x + x - 5 - 25x`

`= 21x  - 6 ->` Bậc `1`.

Lời giải:

$(3x-1)(3x+1)-(x-5)(9x-1)-25x$

$=9x^2-1-(9x^2-46x+5)-25x$

$=21x-6$

Bậc của đa thức sau khi thu gọn là $1$

1: =>y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49

=>24y=20

=>y=56

2: =>y^3+9y^2+27y+27-y^3-3y^2-3y-1=56

=>6y^2+24y-30=0

=>y^2+4y-5=0

=>y=-5; y=1

1: =x^2-2x-8+x^3-6x^2+12x-8+16

=x^3-5x^2+10x

2: =4x^2-8x+x^3-2x+x^3+3x^2+3x+1

=2x^3+7x^2-5x+1

`1)`

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

`= x^3 - 6x^2 + 12x -8`

`2)`

`(y-3)^3= y^3 - 3.y^2 . 3 + 3.y.3^2 - 3^3`

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

`3)`

`(2x-5)^3= (2x)^3 - 3.2x^2 . 5 + 3.2x.5^2 + 5^3`

`= 8x^3 - 12x + 150x + 125`

\(1,\left(x-2\right)^3=x^3-3.x^2.2+3.x.2^2-2^3\\ =x^3-6x^2+12x-8\\ 2,\left(y-3\right)^3=y^3-3.y^2.3+3y.9-27\\ =y^3-9y^2+27y-27\\ 3,\left(2x-5\right)^3=\left(2x\right)^3-3.4x^2.5+3.2x.25-125\\ =8x^3-60x^2+375x-125\)

(1) \(\left(x-2\right)^3=x^3-3x^22+3x2^2-2^3=x^3-6x^2+12x-8\)

(2) \(\left(y-3\right)^3=y^3-3y^23+3y3^2-3^3=y^3-9y^2+27y-27\)

(3) \(\left(2x-5\right)^3=\left(2x\right)^3-3\left(2x\right)^25+3\left(2x\right)5^2-5^3=8x^3-60x^2+150x-125\)

`1)`

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

`= x^3 + 3x^2 + 3x +1`

`2)`

`(x+2)^3=x^3 + 3.x^2 .2 + 3.x.2^2 +2^3`

`= x^3 + 6x^2 + 12x+8`

`3)`

`(x+3)^3=x^3 + 3.x^2 . 3 + 3.x.3^2 + 3^3`

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

`--------------`

`@ ` áp dụng HĐT số 4`

1.(x+1)^3

=x^3+3x^2+3x+1

2.(x+2)^3

=3x+63.

3.(x+3)^3

=3x+9

1.\(\left(x+1\right)^3=x^3+3x^21+3x1^2+1^3=x^3+3x^2+\) \(3x+1.\)

2.\(\left(x+2\right)^3=x^3+3x^22+3x2^2+2^3=x^3+6x^2+\)\(12x+8.\)

3.\(\left(x+3\right)^3=x^3+3x^23+3x3^2+3^3=x^3+9x^2+\)\(27x+27.\)