Bài 1:

a, (3\(x\) + 2)(4\(x\) - 1) - (2\(x\) - 1)(6\(x\) - 5)

 = 12\(x^2\) - 3\(x\) + 8\(x\) - 2 - 12\(x^2\)  + 10\(x\) + 6\(x\) - 5

 = 21\(x\) - 7

b, (2\(x\) + 1)(\(x^2\) - 7y) - 7y(y- 2\(x\) - 1)

= 2\(x^3\) + \(x^2\) - 14\(xy\) - 7y - 7y² + 14\(xy\) + 7y

= 2\(x^3\) + \(x^2\) - 7y²

c, (-4\(x^3\)y⁵ + 2\(xy^2\)) : ( - 5\(xy^2\))

= -2\(xy^2\)( 2\(x^2y^3\) - 1) : ( -5\(xy^2\))

= 0,4.(2\(x^2y^3\) - 1)

= 0,8\(x^2y^3\) - 0,4 

-4\(x^n\)y⁵ ⋮ 7\(x^4\).y\(^n\) ⇔ \(\left\{{}\begin{matrix}x.y⋮7\\n>4\\5>n\end{matrix}\right.\) ⇒  S = \(\varnothing\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`-2x (3x + 2)`

`= -6x^2 - 4x`

`b)`

`2/3x(x^2 - x + 4)`

`= 2/3x^3 - 2/3x^2 + 8/3x`

`c)`

`5ab ( ab - 2a^2b^3)`

`= 5a^2b^2 - 10a^3b^4`

`d)`

`3x(x-5)`

`= 3x^2 - 15x`

A = 12\(x\) - 4\(x^2\) + 3

A = -(4\(x^2\) - 2.2\(x\).3 + 9) + 12

A = -( 2\(x\) - 3)² + 12

    (2\(x\)- 3)² ≥  0 ⇒ -(2\(x\) - 3)² ≤ 0 ⇒- (2\(x\) - 3)² + 12 ≤ 12

Amax = 12⇔ 2\(x\) - 3 =  0 ⇒ \(x\) = \(\dfrac{3}{2}\)

Giá trị lớn nhất của A là 12 xảy ra khi \(x\) = \(\dfrac{3}{2}\)

B = 6\(x\) - \(x^2\) + 3 

B = - (\(x^2\) - 2.3\(x\) + 9) + 12

B = -(\(x\) - 3)² + 12

(\(x\) - 3)² ≥ 0 ⇒ -(\(x\) - 3)² ≤ 0 ⇒ -(\(x\) - 3)² + 12  ≤ 12 

Bmax = 12 ⇔ \(x\) - 3 = 0 ⇒ \(x\) = 3

Giá trị lớn nhất của B là 12 xảy ra khi \(x\) = 3

Bài 1 :

\(A=-x^2+6x+14\)

\(A=-x^2+6x-9+23\)

\(A=-\left(x^2-6x+9\right)+23\)

\(A=-\left(x-3\right)^2+23\)

Vì \(-\left(x-3\right)^2\le0\)

\(\Rightarrow A=-\left(x-3\right)^2+23\le23\)

\(\Rightarrow Max\left(A\right)=23\)

Bài 2 :

\(B=4x^2+12x+30\)

\(\Rightarrow B=4x^2+12x+9+21\)

\(\Rightarrow B=\left(2x+3\right)^2+21\)

Vì \(\left(2x+3\right)^2\ge0\)

\(\Rightarrow B=\left(2x+3\right)^2+21\ge21\)

\(\Rightarrow Min\left(B\right)=21\)

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

a) 25x² - 16

= (5x)² - 4²

= (5x - 4)(5x + 4)

b) 16a² - 9b²

= (4a)² - (3b)²

= (4a - 3b)(4a + 3b)

c) 8x³ + 1

= (2x)³ + 1³

= (2x + 1)(4x² - 2x + 1)

d) 125x³ + 27y³

= (5x)³ + (3y)³

= (5x + 3y)(25x² - 15xy + 9y²)

e) 8x³ - 125

= (2x)³ - 5³

= (2x - 5)(4x² + 10x + 25)

g) 27x³ - y³

= (3x)³ - y³

= (3x - y)(9x² + 3xy + y²)

a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)

b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)

c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)

d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)

g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

a) (x² + 2)²

= (x²)² + 2.x².2 + 2²

= x⁴ + 4x² + 4

b) (x + y + z)²

= [(x + y) + z]²

= (x + y)² + 2(x + y).z + z²

= x² + 2xy + y² + 2xz + 2yz + z²

= x² + y² + z² + 2xy + 2xz + 2yz

sos

a) 25x² - 10xy + y²

= (5x)² - 2.5x.y + y²

= (5x - y)²

b) 4/9 x² + 20/3 xy + + 25y²

= (2/3 x)² + 2.2/3 x.5y + (5y)²

= (2/3 x + 5y)²

c) 9x² - 12x + 4

= (3x)² - 2.3x.2 + 2²

= (3x - 2)²

d) Sửa đề: 16u²v⁴ - 8uv² + 1

= (4uv²)² - 2.4uv².1 + 1²

= (4uv² - 1)²

chỉ cần giải mỗi c và d thui

3\(x^2\) - 4\(x\) - 4 = 0

3(\(x^2\) - 2. \(\dfrac{2}{3}\)\(x\) + \(\dfrac{4}{9}\)) - \(\dfrac{16}{3}\) = 0

3.(\(x-\dfrac{2}{3}\))² = \(\dfrac{16}{3}\)

   (\(x-\dfrac{2}{3}\))² = \(\dfrac{16}{9}\) 

   \(\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{4}{3}\\x-\dfrac{2}{3}=-\dfrac{4}{3}\end{matrix}\right.\)

  \(\left[{}\begin{matrix}x=\dfrac{4}{3}+\dfrac{2}{3}\\x=-\dfrac{4}{3}+\dfrac{2}{3}\end{matrix}\right.\)

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

S = { -\(\dfrac{2}{3}\); 2}

3x² - 4x - 4 = 0

⇔ 3x² - 6x + 2x - 4 = 0

⇔ (3x² - 6x) + (2x - 4) = 0

⇔ 3x(x - 2) + 2(x - 2) = 0

⇔ (x - 2)(3x + 2) = 0

⇔ x - 2 = 0 hoặc 3x + 2 = 0

*) x - 2 = 0

⇔ x = 2

*) 3x + 2 = 0

⇔ 3x = -2

⇔ x = -2/3

Vậy S = {-2/3; 2}

a, y^?  thế em ơi 

y bình phương ...