3x -xy + 6y- 18 = 4 -18

x(3-y) + 6(y -3) = -14 => x(y-3) -6(y-3) =14

(y-3)(x-6) =14

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8 

Bài 2:

\(A=-2x^2+3x-5\)

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

\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)

\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)

Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)

Bài 1:

a)x²-4x²y+4xy

=x(x-4xy+y)

b)đề sai

Bài 3:

3yx + 6x - y = 7

<=> x(3y+6) - (3y+6) = 27

<=> (3y+6)(x+1) = 27

Ta có bảng sau:

Vậy...

x³ - 3x²y + 3xy² - y³ - z³

= (x³ - 3x²y + 3xy² - y³) - z³

= (x - y)³ - z³

= (x - y - z)[(x - y)² + (x - y)z + z²]

= (x - y - z)(x² - 2xy + y² + xz - yz + z³)

--------------------

x² - y² + 8x + 6y + 7

= (x² + 8x + 16) - (y² - 6y + 9)

= (x + 4)² - (y - 3)²

= (x + 4 - y + 3)(x + 4 + y - 3)

= (x - y + 7)(x + y + 1)

a: \(=\left(x^3-3x^2y+3xy^2-y^3\right)-z^3\)

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

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

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

b: \(=x^2+8x+16-y^2+6y-9\)

=(x+4)^2-(y-3)^2

=(x+4+y-3)(x+4-y+3)

=(x+y+1)(x-y+7)

Giúp mình với =(((

`@` `\text {Ans}`

`\downarrow`

`a)`

`(x-1)(y+2)=7`

`=> (x - 1)(y + 2) \in` Ư`(7) = {7; 1; -1; -7}`

Ta có bảng sau:

Vậy, ta có cặp `(x; y)` thỏa mãn `{-1; 8}; {2; 5}; {-9; 0}; {-6; -3}`
`b)`

`x(y - 1) + y = 4`

`=> x(y - 1) + y - 4 = 0`

`=> x(y - 1) + (y - 1) - 3 = 0`

`=> (x + 1)(y - 1) = 3`

`=> (x + 1)(y - 1) \in` Ư`(3) = {-1; -3; 1; 3}`

Ta có bảng sau:

Vậy, ta có cặp `(x; y)` thỏa mãn `{0; 4}; {2; 2}; {-2; -2}; {-4; 0}`