a) x² - 12x - 2x + 24 = 0

<=> x( x - 12 ) - 2( x - 12 ) = 0

<=> ( x - 12 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x-12=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)

b) x² - 5x - 24 = 0

<=> x² + 3x - 8x - 24 = 0

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

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

<=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)

c) 4x² - 12x - 7 = 0

<=> 4x² + 2x - 14x - 7 = 0

<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0

<=> ( 2x + 1 )( 2x - 7 ) = 0

<=> \(\orbr{\begin{cases}2x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

d) x³ + 6x² + 12x + 8 = 0

<=> ( x + 2 )³ = 0

<=> x + 2 = 0

<=> x = -2

e) ( x + 2 )² - x² + 4 = 0

<=> x² + 4x + 4 - x² + 4 = 0

<=> 4x + 8 = 0

<=> 4x = -8

<=> x = -2

f) 2( x + 5 ) = x² + 5x

<=> x² + 5x - 2x - 10 = 0

<=> x( x + 5 ) - 2( x + 5 ) = 0

<=> ( x + 5 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

m) 16( 2x - 3 )² - 25( x - 5 )² = 0

<=> 4²( 2x - 3 )² - 5²( x - 5 )² = 0

<=> [ 4( 2x - 3 ) ]² - [ 5( x - 5 ) ]² = 0

<=> ( 8x - 12 )² - ( 5x - 25 )² = 0

<=> [ 8x - 12 - ( 5x - 25 ) ][ 8x - 12 + ( 5x - 25 ) ] = 0

<=> ( 8x - 12 - 5x + 25 )( 8x - 12 + 5x - 25 ) = 0

<=> ( 3x + 13 )( 13x - 37 ) = 0

<=> \(\orbr{\begin{cases}3x+13=0\\13x-37=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)

n) x² - 6x + 4 = 0

<=> ( x² - 6x + 9 ) - 5 = 0

<=> ( x - 3 )² - ( √5 )² = 0

<=> ( x - 3 - √5 )( x - 3 + √5 ) = 0

<=> \(\orbr{\begin{cases}x-3-\sqrt{5}=0\\x-3+\sqrt{5}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)

a) \(x^2-12x-2x+24=0\)

\(\Leftrightarrow x\left(x-12\right)-2\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)

b) \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x^2+3x\right)-\left(8x+24\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)

c) \(4x^2-12x-7=0\)

\(\Leftrightarrow\left(4x^2-14x\right)+\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)

d) \(x^3+6x^2+12x+8=0\)

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

\(\Rightarrow x=-2\)

e) \(\left(x+2\right)^2-x^2+4=0\)

\(\Leftrightarrow4x+8=0\)

\(\Rightarrow x=-2\)

f) \(2\left(x+5\right)=x^2+5x\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

m) \(16\left(2x-3\right)^2-25\left(x-5\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}8x-12=5x-25\\8x-12=25-5x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=-13\\13x=37\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)

n) \(x^2-6x+4=0\)

\(\Leftrightarrow\left(x-3\right)^2-5=0\)

\(\Leftrightarrow\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)

a, \(x^2-12x-2x+24=0\Leftrightarrow x^2-14x+24=0\Leftrightarrow\left(x-12\right)\left(x-2\right)=0\)

TH1 : x = 12 ; TH2 : x = 2 

b, \(x^2-5x-24=0\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

TH1 : x = 8 ; TH2 : x = -3 

c, \(4x^2-12x-7=0\Leftrightarrow\left(2x+1\right)\left(2x-7\right)=0\)

TH1 : x = -1/2 ; TH2 : x = 7/2

d, \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\Leftrightarrow x=-2\)

Tương tự HĐT thôi :)

a) 5ax - 15ay + 20a = 5a( x - 3y + 4 )

b) 6xy - 12x - 8y = 2( xy - 6x - 4y )

c) 3ab( x - y ) + 3a( y - x ) = 3ab( x - y ) - 3a( x - y ) = ( x - y )( 3ab - 3a ) = 3a( x - y )( b - 1 )

d) x² - xy + 2x - 2y = x( x - y ) + 2( x - y ) = ( x - y )( x + 2 )

e) ax² - 5x² - ax + 5x + a - 5 = x²( a - 5 ) - x( a - 5 ) + ( a - 5 ) = ( a - 5 )( x² - x + 1 )

g) x²y - 4xy² + 4y³ - 36yz² = y( x² - 4xy + 4y² - 36z² ) = y[ ( x² - 4xy + 4y² ) - 36z² ] = y[ ( x - 2y )² - ( 6z )² ] = y( x - 2y - 6z )( x - 2y + 6z )

h) 4xy - x² - 4y² + m² - 6m + 9

= ( m² - 6x + 9 ) - ( x² - 4xy + 4y² )

= ( m - 3 )² - ( x - 2y )²

= ( m - 3 - x + 2y )( m - 3 + x - 2y )

i) x² + x - 12 = x³ - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )

k) 5x² + 14x - 3 = 5x² - x + 15x - 3 = x( 5x - 1 ) + 3( 5x - 1 ) = ( 5x - 1 )( x + 3 )

m) x² - 5xy + 4y² = x² - xy - 4xy + 4y² = x( x - y ) - 4y( x - y ) = ( x - y )( x - 4y ) < đã sửa đề >

n) 3x² - 5xy + 2y² + 4x - 4y = ( 3x² - 5xy + 2y² ) + ( 4x - 4y ) = ( 3x² - 3xy - 2xy + 2y² ) + 4( x - y ) = [ 3x( x - y ) - 2y( x - y ) ] + 4( x - y ) = ( x - y )( 3x - 2y ) + 4( x - y ) = ( x - y )( 3x - 2y + 4 )

f) 2x³ + 4x²y + 2xy² = 2x( x² + 2xy + y² ) = 2x( x + y )²

a, \(5ax-15ay+20a=5a\left(x-5y+4\right)\)

b, sai 

c, \(3ab\left(x+y\right)+3a\left(y-x\right)=3ab\left(x+y\right)-3a\left(x+y\right)=\left(3ab-3a\right)\left(x+y\right)\)

d, \(x^2-xy+2x-2y=x\left(x+2\right)-y\left(x+2\right)=\left(x-y\right)\left(x+2\right)\)

Tượng tự ... 

a) \(5ax-15ay+20a\)

\(=5a\left(x-3y+4\right)\)

b) \(6xy-12x-8y\)

\(=6\left(xy-2x-3y\right)\)

c) \(3ab\left(x-y\right)+3a\left(y-x\right)\)

\(=3a\left(x-y\right)\left(b-1\right)\)

d) \(x^2-xy+2x-2y\)

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

e) \(ax^2-5x^2-ax+5x+a-5\)

\(=\left(a-5\right)\left(x^2-x+1\right)\)

3x² + 9x - 30 

= 3 ( x² + 3x - 10 )

= 3 ( x² + 5x - 2x - 10 ) 

= 3 [ x ( x + 5 ) - 2 ( x + 5 ) ] 

= 3 ( x + 5 ) ( x - 2 ) 

3x² + 9x - 30

= 3 ( x² + 3x - 10 )

= 3 [ ( x² - 2x ) + ( 5x - 10 ) ]

= 3 [ x ( x - 2 ) + 5 ( x - 2 ) ]

= 3 ( x + 5 ) ( x - 2 )