a: \(\left(3x+4y\right)\left(9x^2-12y+16y^2\right)\)

\(=27x^3-36xy+48xy^2+36x^2y-48y^2+64y^3\)

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

\(=x^3+9x^2+27x+27-\left(9x^2-6x+1\right)\)

\(=x^3+9x^2+27x+27-9x^2+6x-1\)

\(=x^3+33x+26\)

`#3107.101107`

`1.`

`a,`

`(3x + 4y)(9x^2 - 12xy + 16y^2)?`

`= (3x)^3 + (4y)^3`

`= 27x^3 + 64y^3`

`b,`

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

`= x^3 + 9x^2 + 27x + 27 - (9x^2 - 6x + 1)`

`= x^3 + 9x^2 + 27x + 27 - 9x^2 + 6x - 1`

`= x^3 + 33x + 26`

_____

Sử dụng HĐT:

`A^3 + B^3 = (A + B)(A^2 + AB + B^2)`

`(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`

`(A - B)^2 = A^2 - 2AB + B^2.`

Đề lỗi rồi kìa, bạn viết lại đi tridung

A= (\(\dfrac{1}{2}\)+\(\dfrac{3}{4}\)-\(\dfrac{4}{4}\))X²Y² = \(\dfrac{1}{4}\)x²y²

A = 3x^3 +6x^2 + 3xy^3

x= 1 phần 2 ;  p = -1 phần 3

A=3.1 phần 2^3 . -1 phần 3     + 6.(1 phần 2)^2 . (-1 Phần 3)^2+3 1 phần 2 . (-1 phần 3)^3

=-1 phần 8      + -1 phần 2 - 1 phần 2

= -1 phần 4

a) Ta có: \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)

\(=\left[\left(3-xy^2\right)-\left(2+xy^2\right)\right]\cdot\left[\left(3-xy^2\right)+\left(2+xy^2\right)\right]\)

\(=\left(3-xy^2-2-xy^2\right)\cdot\left(3-xy^2+2+xy^2\right)\)

\(=5\cdot\left(1-2xy^2\right)\)

\(=5-10xy^2\)

b) Ta có: \(9x^2-\left(3x-4\right)^2\)

\(=\left[3x-\left(3x-4\right)\right]\left[3x+\left(3x-4\right)\right]\)

\(=\left(3x-3x+4\right)\cdot\left(3x+3x-4\right)\)

\(=4\cdot\left(6x-4\right)\)

\(=24x-16\)

c) Ta có: \(\left(a-b^2\right)\left(a+b^2\right)\)

\(=a^2-b^4\)

d) Ta có: \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

\(=\left(a^2+2a\right)^2-9\)

\(=a^4+4a^3+4a^2-9\)

e) Ta có: \(\left(x-y+6\right)\left(x+y-6\right)\)

\(=x^2+xy-6x-yx-y^2+6y+6x+6y-36\)

\(=x^2-y^2+12y-36\)

f) Ta có: \(\left(y+2z-3\right)\left(y-2z-3\right)\)

\(=\left(y-3\right)^2-\left(2z\right)^2\)

\(=y^2-6y+9-4z^2\)

g) Ta có: \(\left(2y-5\right)\left(4y^2+10y+25\right)\)

\(=\left(2y\right)^3-5^3\)

\(=8y^3-125\)

h) Ta có: \(\left(3y+4\right)\left(9y^2-12y+16\right)\)

\(=\left(3y\right)^3+4^3\)

\(=27y^3+64\)

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

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

\(=x^3-9x^2+27x-27-\left(x^3-6x^2+12x-8\right)\)

\(=x^3-9x^2+27x-27-x^3+6x^2-12x+8\)

\(=-3x^2+15x-19\)

j) Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=\left[\left(x+y\right)-\left(x-y\right)\right]\cdot\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

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

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

1.

a. \((x+1)(x^2-x+1)-(x-1)(x^2+x+1)\)

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

b.

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

2.

a. \(x^2-4y^2+12y-9=x^2-\left[\left(2y\right)^2-2\cdot2y\cdot3+3^2\right]=x^2-\left(2y-3\right)^2=\left(x-2y+3\right)\left(x+2y-3\right)\)

b.

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

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

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

\(=\left(x+y\right)\left[3\left(x+y\right)+5\left(x-y\right)\right]\)

\(=\left(x+y\right)\left(3x+3y+5x-5y\right)\)

\(=\left(x+y\right)\left(8x-2y\right)=2\left(x+y\right)\left(4x-y\right)\)

3.

a.

\(x^3-x=0\)

\(\Leftrightarrow x\left(x^2-1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

b.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+1\\3x+1=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)