1) a) 2a + 2b = 2(a + b) ; 

b) 9a - 9b = 9(a - b) ; 

c) 3a - 6b - 9c = 3(a - 2b - 3c) 

d) ab - ac = a(b - c) 

e) 5a - 10ax - 15a = -10a - 10ax = 10a(x + 1) 

f) 3a(ax - 2ay + 4) 

g) 5a²(x - y) + 10a(x - y) 

= 5a(x - y)(a + 2) 

2) ax + ay + bx + by 

 = a(x + y) + b(x + y) 

= (a + b)(x + y) 

b) a² - 49 = (a - 7)(a  + 7) 

c) 9a² - 1 = (3a - 1)(3a + 1) 

d) \(\frac{1}{4}a^2-b^2=\left(\frac{1}{2}a-b\right)\left(\frac{1}{2}a+b\right)\)

e) x² + 14x + 49 = (x + 7)²

f) 4x² + 20x + 25 = (2x + 5)²

g) 4x⁴ + 20x² + 25 = (2x² + 5)²

h) 2x³ + 8x² + 8x = 2x(x²  + 4x + 4) = 2x(x + 2)²

i) 2x³ + 16 = 2(x³ + 8) = 2(x + 2)(x² - 2x + 4) 

3) x² + 4x + 3 = x² + x + 3x + 3 = x(x + 1) + 3(x + 1) = (x + 1)(x + 3) 

x² + 7x + 10 = x² + 2x + 5x + 10 = x(x + 2) + 5(x + 2) = (x + 2)(x + 5) 

fuk                                                  hihi

ta có 2(x^2+x+1)/x^2+1

=2x^2+2x+2/x^2+1

=1+(x+1)^2/x^2+1>=1 với mọi x

dấu bằng xảy ra khi x=-1

bạn tự kết luận nha

= \(\left(x^2-9y^2\right)\)

\(^{x^2}\)-\(9y^2\)

x³ - 9x - 8x - 24 = 0 

<=> x(x² - 9) - 8(x + 3) = 0

<=> x(x - 3)(x + 3) - 8(x + 3) = 0

<=> (x + 3)(x² - 3x - 8) = 0 

<=> \(\left(x+3\right)\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{41}{4}\right)=0\)

<=> \(\left(x+3\right)\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{41}}{2}\right)^2\right]=0\)

<=> \(\left(x+3\right)\left(x-\frac{3}{2}+\frac{\sqrt{41}}{2}\right)\left(x-\frac{3}{2}-\frac{\sqrt{41}}{2}\right)=0\)

<=> \(\orbr{\begin{cases}x=-3\\x=\frac{3\pm\sqrt{41}}{2}\end{cases}}\)

(x+2y)^2=x^2+2.x.2y+(2y)^2=x^2+4xy+4y^2

thank, đáp án đứng r

a) A = ( x - 2y )³ + ( x + 2y )³ - 2x ( x² + y )=

       = x³ - 6x²y + 12xy² - 8y³ + x³ + 6x²y + 12xy² + 8y³ - 2x³ - 2xy

       =                       24xy² - 2xy

b) B = ( x - 1 )( x + y ) ( x - y )  - x²( x - 1 )= 

       =  ( x -1 )( x² - y² ) - x² ( x - 1 )

       = ( x - 1 )( x² - y² - x² )

       = -y² ( x - 1 ) 

c ) C =  ( x + 2)² - 2( x + 2 )( x - 8 ) + ( x - 8 ) ² = 

        = ( x + 2 - x + 8 ) ² 

        = 10²

          = 100

HOk tốt!!!!!!!!!! 

Bài 2 : a) \(2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}}\)

b) \(2\left(2x-1\right)+6x\left(2x-1\right)=0\Leftrightarrow\left(2x-1\right)\left(2+6x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2+6x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\6x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{6}=-\frac{1}{3}\end{cases}}}\)

c) \(\left(x-3\right)^2-\left(2x+6\right)^2=0\Leftrightarrow\left(x-3-2x-6\right)\left(x-3+2x+6\right)=0\)

\(\Leftrightarrow\left(-x-9\right)\left(3x+3\right)=0\Leftrightarrow\orbr{\begin{cases}-x-9=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=-1\end{cases}}}\)

Tự KL cho các phần