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15 tháng 12 2016

\(2x^3+x^2-4x-12\)

\(=\left(2x^3-4x^2\right)+\left(5x^2-10x\right)+\left(6x-12\right)\)

\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(2x^2+5x+6\right)\)

15 tháng 12 2016

Đặt 2x ra. là oki nà

22 tháng 11 2016

a,

(x^2+x)^2+4x^2+4x-12

=x^4 + 2x^3 + 5x^2 + 4x -12

=(x-1)(x^3+3x^2+8x+12)

=(x-1)(x+2)(x^2+x+6)

 

22 tháng 11 2016

b , 3x^2+6xy+3y^2-12

=3(x^2+2xy+y^2-4)

=3[(x+y)^2 -2^2]

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

28 tháng 7 2016

1.  x^3-19x-30 
=x^3-25x+6x-30 
=x(x^2-25)+6(x-5) 
=x(x+5)(x-5)+6(x-5) 
=(x-5)(x^2+5x+6) 
=(x-5)(x^2+2x+3x+6) 
=(x-5)[x(x+2)+3(x+2)] 
=(x-5)(x+2)(x+3)

28 tháng 7 2016

 2.

a + b + c = 0 
<=> (a + b + c)² = 0 
<=> a² + b² + c² + 2(ab + bc + ca) = 0 
<=> a² + b² + c² = -2(ab + bc + ca) ------------(1) 

CẦn chứng minh: 

2(a^4 + b^4 + c^4) = (a² + b² + c²)² 

<=> 2(a^4 + b^4 + c^4) = a^4 + b^4 + c^4 + 2(a²b² + b²c² + c²a²) 

<=> a^4 + b^4 + c^4 = 2(a²b² + b²c² + c²a²) 

<=> (a² + b² + c²)² = 4(a²b² + b²c² + c²a²) ---(cộng 2 vế cho 2(a²b² + b²c² + c²a²) ) 

<=> [-2(ab + bc + ca)]² = 4(a²b² + b²c² + c²a²) ----(do (1)) 

<=> 4.(a²b² + b²c² + c²a²) + 8.(ab²c + bc²a + a²bc) = 4(a²b² + b²c² + c²a²) 

<=> 8.(ab²c + bc²a + a²bc) = 0 

<=> 8abc.(a + b + c) = 0 

<=> 0 = 0 (đúng), Vì a + b + c = 0 

=> Đpcm

27 tháng 9 2019

\(G=2x^2-3x+1=2x^2-2x-x+1\)

\(=2x\left(x-1\right)-\left(x-1\right)=\left(2x-1\right)\left(x-1\right)\)

\(H=-x^2+5x-4=-x^2+4x+x-4\)

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

\(I=x^2+4x+3=x^2+3x+x+3\)

\(=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)

\(K=2x^2+7x+5=2x^2+2x+5x+5\)

\(=2x\left(x+1\right)+5\left(x+1\right)=\left(2x+5\right)\left(x+1\right)\)

\(L=-3x^2-5x-2=-3x^2-3x-2x-2\)

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

27 tháng 9 2019

G = 2x2 - 3x +1 =  2x2 -2x -x +1 =(x-1).(2x-1)

H = -x2 + 5x - 4 = -x2 + 4x +x-4 = (x-4).(1-x)

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

K = 2x2 + 7x + 5 = 2x2 + 2x + 5x + 5 = (x+1).(2x+5)

L = -3x2 -5x -2 = -3x2 - 3x - 2x - 2 = -3.x(x+1) - 2.(x+1) = (x+1).(-3x-2)

31 tháng 7 2016

a) (2x - 1)2 - (x + 3)2

= (2x - 1 - x - 3).(2x - 1 + x + 3)

= (x - 4).(3x + 2)

b) x2.(x - 3) + 12 - 4x

= x2.(x - 3) - 4x + 12

= x2.(x - 3) - 4.(x - 3)

= (x - 3).(x2 - 4)

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

28 tháng 9 2018

Áp dụng HĐT:

    a2 - b2 = (a - b)(a + b)

\(\left(2x-1\right)^2-\left(x+3\right)^2\)

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

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

22 tháng 9 2020

a) x2 + x - 12 = x2 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )

b) x2 - 4x - 5 = x2 + x - 5x - 5 = x( x + 1 ) - 5( x + 1 ) = ( x + 1 )( x - 5 )

c) x2 - 2x - 3 = x2 + x - 3x - 3 = x( x + 1 ) - 3( x + 1 ) = ( x + 1 )( x - 3 )

d) x2 - 2x - 8 = x2 + 2x - 4x - 8 = x( x + 2 ) - 4( x + 2 ) = ( x + 2 )( x - 4 )

e) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )

f) x2 - 6x + 8 = x- 2x - 4x + 8 = x( x - 2 ) - 4( x - 2 ) = ( x - 2 )( x - 4 )

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

h) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )

i) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )

j) x2 - 5x - 14 = x2 + 2x - 7x - 14 = x( x + 2 ) - 7( x + 2 ) = ( x + 2 )( x - 7 )

25 tháng 6 2016

bạn áp dụng các hằng đẳng thức là Ok:

a) \(-4x^2+12xy-9y^2+25=-\left(2x\right)^2+2.2x.3y-\left(3y\right)^2+25\)

\(-\left(2x-3y\right)^2+25=\left(5-2x+3y\right)\left(5+2x-3y\right)\)

b) \(x^3-3x^2+3x-1=\left(x^3-1\right)-\left(3x^2-3x\right)=\left(x-1\right)\left(x^2+x+1^2\right)-3x\left(x-1\right)\)

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

c) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a^2-1\right)+2\left(a+1\right)\right]\)

\(a^2\left[a^2\left(a+1\right)\left(a-1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3+a^2+2\right)\)

25 tháng 6 2016

Thanks bạn hiền hiuhiu

3 tháng 7 2018

a/ 9(x-y)2 - 4(x+y)2

= [3(x-y)]2 - [2(x+y)]2

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

= (x-5y)(5x-y)

b/ x3 + 1 - x2 - x

= (x + 1)(x2 + x + 1) - x(x+1)

= (x+1)(x2 + x + 1 - x)

= (x+1)(x2 + 1)

c/ x2 - 2x - 4y2 - 4y

= (x2 - 4y2) - (2x + 4y)

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

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

4 tháng 8 2016

a)\(x^2+10x=24\)

\(\Leftrightarrow x^2+10x-24=0\)

\(\Leftrightarrow x^2-2x+12x-24=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)

b)\(4x^2+4x=24\)

\(\Leftrightarrow4x^2+4x-24=0\)

\(\Leftrightarrow4\left(x^2+x-6\right)=0\)

\(\Leftrightarrow x^2+x-6=0\)

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)

c)\(4x^2-4x=48\)

\(\Leftrightarrow4x^2-4x-48=0\)

\(\Leftrightarrow4\left(x^2-x-12\right)=0\)

\(\Leftrightarrow x^2-x-12=0\)

\(\Leftrightarrow x^2+3x-4x-12=0\)

\(\Leftrightarrow x\left(x+3\right)-4\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

4 tháng 8 2016

\(a,x^2+10x=24\)
\(\Leftrightarrow x^2+10x-24=0\)
\(\Leftrightarrow x^2-2x+12x-24=0\)
\(\Leftrightarrow x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-12 }\)
\(b,4x^2+4x=24\)
\(\Leftrightarrow4x^2+4x-24=0\)
\(\Leftrightarrow4x^2-8x+12x-24=0\)
\(=4x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
Vậy hoặc \(\text{Vậy x=2 hoặc x=-3 }\)
\(c,4x^2-4x=48\)
\(\Leftrightarrow4x^2-4x-48=0\)
\(\Leftrightarrow\left[\left(2x\right)^2-2.2x+1^2\right]-1^2-48=0\)
\(\Leftrightarrow\left(2x-1\right)^2-49=0\)
\(\Leftrightarrow\left(2x-1\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-1-7\right)\left(2x-1+7\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-8=0\\2x+6=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)
\(\text{Vậy x=4 hoặc x=-3 }\)