`B = -x^2 + 6x - 2`

`= -x^2 + 6x - 9 + 7`

`= -(x^2 - 2.3x + 3^2) + 7`

`= -(x - 3)^2 + 7`

Do  `-(x - 3)^2 <=0` với mọi x

`=> -(x - 3)^2 + 7 <= 7` với mọi x

Dấu = có khi: 

`x - 3= 0 `

`<=> x = 3`

Vậy ...

Diện tích xung quanh của phần chụp đèn là:
Sxq = 1/2 C. d = 1/2 * (4*20) * 22,4 = 896 (cm^2)

a; A =  5\(x^2\) + 10\(xy\) - 4\(x\) - 8y

   A = (5\(x^2\) + 10\(xy\)) - (4\(x\) - 8y) 

  A = 5\(x\).(\(x\) + 2\(y\)) - 4.(\(x+2y\))

 A =  (\(x+2y\)).(5\(x\) - 4)

B = 4\(x^2\) + 4\(x\) - y² + 1

B = (4\(x^2\) + 4\(x\) + 1) - y²

B = [(2\(x\))² + 2.2\(x\).1 + 1²] - y²

B = [2\(x\) + 1]² - y²

B = (2\(x+1\) - y)(2\(x+1\) - y)

giúp mình với

Chọn C nhé

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

\(\Leftrightarrow\left(4x-1\right)\left(4x+1\right)-\left(4x-1\right)^2=0\)

\(\Leftrightarrow\left(4x-1\right)\left[\left(4x+1\right)-\left(4x-1\right)\right]=0\)

\(\Leftrightarrow2\left(4x-1\right)=0\)

\(\Leftrightarrow4x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

(4\(x\)  - 1).(4\(x\) + 1) = (4\(x\) - 1)²

(4\(x-1\)).(4\(x\) + 1) - (4\(x\) - 1)² = 0

(4\(x\) - 1).(4\(x\) + 1 - 4\(x\) + 1) = 0

 (4\(x\) - 1).[(4\(x\) - 4\(x\)) + (1 +1)] = 0

(4\(x\) - 1).[0 + 2] = 0

(4\(x\) - 1).2 = 0

 4\(x\) - 1 = 0 

 4\(x\) = 1

   \(x=\dfrac{1}{4}\)

Vậy \(x=\dfrac{1}{4}\) 

`x^3 - y^6 `

`= x^3 - (y^2)^3`

`= (x - y^2) (x^2 + xy^2 + y^4) `

He's really badly-dressed 

A. What

B nhé

hỏi về phương tiện di chuyển ta How

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