Ủa đề bài là sao mình không hiều bạn?

Làm hai trường hợp nhé :v trúng cái nào thì trúng

1. x³ - x = 0

<=> x( x² - 1 ) = 0

<=> x( x - 1 )( x + 1 ) = 0

<=> x = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

<=> x = 0 hoặc x = ±1

2. x³ + x = 0

<=> x( x² + 1 ) = 0

<=> x = 0 hoặc x² + 1 = 0

<=> x = 0 < vì x² + 1 ≥ 1 > 0 ∀ x >

Ta có: \(2ax^3+6ax^2+6ax+18a\)

\(=2\left[\left(ax^3+3ax^2\right)+\left(3ax+9a\right)\right]\)

\(=2a\left[x^2\left(x+3\right)+3\left(x+3\right)\right]\)

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

2ax³ + 6ax² + 6ax + 18a

= 2a( x³ + 3x² + 3x + 9 )

= 2a[ ( x³ + 3x² ) + ( 3x + 9 ) ] 

= 2a[ x²( x + 3 ) + 3( x + 3 ) ]

= 2a( x + 3 )( x² + 3 )

mik mới lớp 6 nên hok bik làm :))

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

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

\(=z^2\)

\(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2-2x+1-25=0\)

\(\Leftrightarrow\left(x-1\right)^2-25=0\)

\(\Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

Vậy \(x=6\)hoặc \(x=-4\)

x² - 2x = 24

<=> x² - 2x - 24 = 0

<=> x² - 2x + 1 - 25 = 0

<=> (x - 1)² - 25 = 0

<=> (x - 1)² - 5² = 0

<=> [(x - 1) + 5] [(x - 1) - 5] = 0

<=> (x - 1 + 5) (x - 1 - 5) = 0

<=> (x + 4) (x - 6) = 0

=> x + 4 = 0 hoặc x - 6 = 0

     x + 4 = 0 => x = -4

     x - 6 = 0 => x = 6

Vậy x = -4 hoặc x = 6

#Học tốt!!

~NTTH~

Xét biểu thức \(\frac{14x^2-8x+9}{3x^2+6x+9}=\frac{14x^2-8x+9}{3x^2+6x+9}-\frac{2}{3}+\frac{2}{3}=\frac{3\left(14x^2-8x+9\right)-2\left(3x^2+6x+9\right)}{3\left(3x^2+6x+9\right)}+\frac{2}{3}=\frac{36x^2-36x+9}{3\left(3x^2+6x+9\right)}+\frac{2}{3}=\frac{\left(6x-3\right)^2}{3\left(3x^2+6x+9\right)}+\frac{2}{3}\ge\frac{2}{3}\)Đẳng thức xảy ra khi x = ¹/₂

Ta có: \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)

\(=\left[\left(x-2\right)\left(x-5\right)\right]\cdot\left[\left(x-3\right)\left(x-4\right)\right]+1\)

\(=\left(x^2-7x+10\right)\cdot\left(x^2-7x+12\right)+1\)

\(=\left[\left(x^2-7x+11\right)-1\right]\cdot\left[\left(x^2-7x+11\right)+1\right]\)

\(=\left(x^2-7x+11\right)^2-1+1\)

\(=\left(x^2-7x+11\right)^2\)

\(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)   

\(=\left(x-2\right)\left(x-5\right)\left(x-4\right)\left(x-3\right)+1\)   

\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)   

Đặt t = \(x^2-7x\)   

\(t\left(t+2\right)+1\)   

\(=t^2+2t+1\)   

\(=\left(t+1\right)^2\)   

\(=\left(x^2-7x+1\right)^2\)

PT đa thức thành nhân tử?

Ta có: \(4x\left(x-y\right)+3\left(y-x\right)^2\)

\(=4x\left(x-y\right)+3\left(x-y\right)^2\)

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

\(=\left(x-y\right)\left(7x-3y\right)\)

\(4x\left(x-y\right)+3\left(y-x\right)^2\)

\(=4x\left(x-y\right)-3\left(x-y\right)^2\)

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

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

\(=\left(x-y\right)\left(7x+3y\right)\)

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

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

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