a) x²+ 2xy + y²- x - y

= ( x + y )² - ( x + y )

= ( x + y )( x + y - 1 )

b) 2x³ + 6x² = 12x + 8

= 2 ( x³ + 3x² + 6x + 4 )

= (2x + 2 )( x + 2x + 4 )

a) x-3= (3x)²=(x-3)-(x-3)²=0

(x-3)-(x-3)²=0

(x-3).(4-x)=0

 x-3 = 0 ⇒x =3

 4-x = 0 ⇒x  =4

vậy x = 3 hoặc 4

b)x³ + \(\dfrac{3}{2}\)x² + \(\dfrac{3}{4}\) + \(\dfrac{1}{8}\) =  \(\dfrac{1}{64}\)

(x +\(\dfrac{1}{2}\))⁸=(\(\dfrac{1}{4}\))³

x + \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\)

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

vậy x = -\(\dfrac{1}{4}\)

do AB//CD( vì cùng vuông govs với BD)

nên áp dụng định lí ta-lét,ta có 

EB//ED=AB//CD

⇒EB/6= 150/4

⇒EB =150.6/4=225 (cm)

a) do d // CD,mà M,N,P ϵ d nên MP // CD,PN//CD,MN//CD

Do ABCD là hình thang nên AB//CD,do đóPN//AB

xét ΔACD với MP//CD, ta có\(\dfrac{AM}{MD}\)=\(\dfrac{AP}{PC}\)(định lí ta-lét)(1)

xét ΔABC với PN//AB,ta có\(\dfrac{AP}{PC}\)=\(\dfrac{BN}{NC}\)(định lí ta-lét)(2)

TỪ (1) và (2) ⇒=\(\dfrac{BN}{NC}\)(=\(\dfrac{AP}{PC}\)\(\dfrac{AM}{MD}\))

b) vì MD=2MA nên \(\dfrac{MA}{MD}\)=\(\dfrac{1}{2}\)⇒\(\dfrac{AM}{AD}\)=\(\dfrac{1}{3}\)

xét ΔADC với MP//CD có\(\dfrac{AM}{AD}\)=\(\dfrac{MP}{DC}\)(định lí ta-lét)

⇒\(\dfrac{MP}{DC}\)=\(\dfrac{1}{3}\)⇒MP=\(\dfrac{1}{3}\)DC=2cm

vì \(\dfrac{AM}{DC}\)=\(\dfrac{1}{3}\)⇒\(\dfrac{AP}{AC}\)=\(\dfrac{1}{3}\)⇒\(\dfrac{PC}{CA}\)=\(\dfrac{2}{3}\)

xét ΔABC với PN//AB có \(\dfrac{CP}{CA}\)=\(\dfrac{PN}{AB}\)(định lí ta lét)

⇒\(\dfrac{PN}{AB}\)=\(\dfrac{2}{3}\)⇒PN=\(\dfrac{2}{3}\)AB=\(\dfrac{8}{3}\)cm

mà MN=MP+PM=2+\(\dfrac{8}{3}\)=\(\dfrac{14}{3}\)cm

a) 3x(x - 1) - (1+x) = 0

⇔3x(x - 1) - (1+x) = 0

$\iff$(x – 1)(3x + 1) = 0

$\iff [$$\begin{array}{c}x-1=0\\ 3x+1=0\end{array}\iff$$[$x=1x=\(\dfrac{-1}{3}\)

b)x²-9x=0

⇔x(x-9)=0

⇔x=0 hoặc x-9=0

a) x²+25-10x

= (x²-10x+25)

= (x²-10x+5²)

= (x - 5)²

b)-8y³+x³ 

= -8y³+2.-8y.x+x³