a: \(x^3+8y^3-\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

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

=0

b: \(\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)+8y^3\)

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

\(=27x^3-8y^3+8y^3=27x^3\)

c: \(\left(x+2\right)^3-\left(x-2\right)^3-12\left(x^2+1\right)\)

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

=4

d: \(\left(2x-1\right)\left(4x^2+2x+1\right)-8\left(x+2\right)\left(x^2-2x+4\right)\)

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

\(=8x^3-1-8x^3-64=-65\)

a) $x^3+8y^3-(x+2y)(x^2-2xy+4y^2)$

$=x^3+8y^3-(x+2y)[x^2-x.2y+(2y)^2]$

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

$=0$

b) $(3x-2y)(9x^2+6xy+4y^2)+8y^3$

$=(3x-2y)[(3x)^2+3x.2y+(2y)^2]+8y^3$

$=(3x)^3-(2y)^3+8y^3$

$=27x^3-8y^3+8y^3=27x^3$

c) $(x+2)^3-(x-2)^3-12(1+x^2)$

$=[(x+2)-(x-2)][(x+2)^2+(x+2)(x-2)+(x-2)^2]-12-12x^2$

$=(x+2-x+2)(x^2+4x+4+x^2-4+x^2-4x+4)-12-12x^2$

$=4(3x^2+4)-12-12x^2$

$=12x^2+16-12-12x^2$

$=4$

d) $(2x-1)(4x^2+2x+1)-8(x+2)(x^2-2x+4)$

$=(2x-1)[(2x)^2+2x.1+1^2]-8(x+2)(x^2-x.2+2^2)$

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

$=8x^3-1-8(x^3+8)$

$=8x^3-1-8x^3-64$

$=-64$

#$\mathtt{Toru}$

\(2x^3-5x-6\\ =2x^3-4x^2+4x^2-8x+3x-6\\ =2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\\=\left(2x^2+4x+3\right)\left(x-2\right)\)

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

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

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

Em ghi là đường  cao H là sai, phải ghi là BH mới đúng vì vậy Olm bảo em làm  sai em hiểu  chưa nhỉ?

\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\\ =x^2y-x^2z+y^2z-xy^2+xz^2-yz^2\\ =\left(x^2y-x^2z\right)+\left(y^2z-yz^2\right)-\left(xy^2-xz^2\right)\\ =x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y^2-z^2\right)\\ =x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y+z\right)\left(y-z\right)\\ =\left(y-z\right)\left[x^2+yz-x\left(y+z\right)\right]\\ =\left(y-z\right)\left(x^2+yz-xy-xz\right)\\ =\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]\\ =\left(y-z\right)\left(x-z\right)\left(x-y\right)\)

\(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{7x-1}{4x+3}-\dfrac{x}{x-5}\left(x\ne-\dfrac{3}{4};x\ne5\right)\\ \Leftrightarrow\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{\left(7x-1\right)\left(x-5\right)}{\left(4x+3\right)\left(x-5\right)}-\dfrac{x\left(4x+3\right)}{\left(4x+3\right)\left(x-5\right)}\\ \Leftrightarrow x^3-\left(x-1\right)^3=\left(7x-1\right)\left(x-5\right)-x\left(4x+3\right)\\ \Leftrightarrow x^3-x^3+3x^2-3x+1=7x^2-35x-x+5-4x^2-3x\\ \Leftrightarrow3x^2-3x+1=3x^2-39x+5\\ \Leftrightarrow-3x+1=-39x+5\\ \Leftrightarrow-3x+39x=5-1\\ \Leftrightarrow36x=4\\ \Leftrightarrow x=\dfrac{4}{36}=\dfrac{1}{9}\left(tm\right)\)

\(2\left(x-1\right)^2-3\left(x+3\right)^2+\left(2x-1\right)\left(2x+1\right)\\ =2\left(x^2-2x+1\right)-3\left(x^2+6x+9\right)+\left[\left(2x\right)^2-1^2\right]\\ =2x^2-4x+2-3x^2-18x-27+4x^2-1\\ =\left(2x^2-3x^2+4x^2\right)+\left(-4x-18x\right)+\left(2-27-1\right)\\ =3x^2-22x-26\)

(d1) 2x + y = 3 => y = -2x + 3

(d2) x - y = 6 => y = x - 6 

Ta có pt tọa độ giao điểm của (d1) và (d2) là:

- 2x + 3 = x - 6

=> x + 2x = 3 + 6

=> 3x = 9

=> x = 9 : 3 = 3 

Thay x = 3 vào (d2) ta có: y = 3 - 6 = -3

Tọa độ giao điểm là: (3;-3)

`5x+4x^2=0`

`<=>x(5+4x)=0`

TH1: x=0

TH2: `5+4x=0`

`<=>4x=-5`

`<=>x=-5/4`

Vậy: ... 

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

\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)

\(=x^3+4^3=x^3+64\)

b: \(\left(x-5\right)\left(x^2+5x+25\right)\)

\(=\left(x-5\right)\left(x^2+x\cdot5+5^2\right)\)

\(=x^3-5^3=x^3-125\)

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

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

\(=x^6-\dfrac{1}{27}\)

d: \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)\)

\(=8x^3-4x^2y+4x^2y-2xy^2+2xy^2-y^3\)

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

\(a.\left(x+4\right)\left(x^2-4x+16\right)\\ =\left(x+4\right)\left(x^2-4\cdot x+4^2\right)\\ =x^3+4^3\\ =x^3+64\\ b.\left(x-5\right)\left(x^2+5x+25\right)\\ =\left(x-5\right)\left(x^2+5\cdot x+5^2\right)\\ =x^3-5^3\\ =x^3-125\\ c.\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\\ =\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}\cdot x^2+\left(\dfrac{1}{3}\right)^2\right]\\ =\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\\ =x^6-\dfrac{1}{27}\\ d.\left(4x^2+2xy+y^2\right)\left(2x-y\right)\\ =\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\\=\left(2x\right)^3-y^3\\ =8x^3-y^3\)