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N
23 tháng 7 2017
a) x2 ( x - 3 ) + 12 - 4x = x2 ( x - 3 ) + 4(3 - x) = x2 ( x - 3 ) - 4(x - 3) = (x - 3) (x2 -4) = (x-3) (x-2 ) (x+2)
b) x2 - 4 + (x+2)2 = (x- 2) (x+2) + (x+2)2 = (x+2)(x -2 + x + 2) = 2x(x+ 2)
c) x3 - 2x2 + x - xy2 = x( x2 - 2x + 1 - y2) = x [ (x-1)2 - y2 ] = x(x- 1- y) (x-1 + y)
d) x3 - 4x2 - 12x + 27 = x3 + 3x2 - 7x2 -21x + 9x + 27
= (x3 + 3x2) - (7x2 + 21x) + (9x + 27)
= x2(x + 3) - 7x(x + 3) + 9(x + 3)
= (x + 3)(x2 - 7x + 9)
NB
1
4 tháng 8 2017
=x4-4x3-2x2+12x+9
=x4+x3-5x3-5x2+3x2+3x+9x+9
=x3(x+1)-5x2(x+1)+3x(x+1)+9(x+1)
=(x+1)(x3-5x2+3x+9)
=(x+1)(x3+x2-6x2-6x+9x+9)
=(x+1)(x2(x+1)-6x(x+1)+9(x+1))
=(x+1)(x+1)(x2-6x+9)
=(x+1)2(x-3)2
29 tháng 6 2017
Ta có : x3 + 2x2 + 2x + 1
= x3 + x2 + (x2 + 2x + 1)
= x2(x + 1) + (x + 1)2
= (x + 1) ( x2 + x + 1)
29 tháng 6 2017
a)\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2\)
\(=a^2b^2+a^2+b^2+1\)
\(=a^2\left(b^2+1\right)+\left(b^2+1\right)\)
\(\left(a^2+1\right)\left(b^2+1\right)\)
b)\(x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
\(=\left(x^2+x+1\right)\left(x+1\right)\)
4 tháng 8 2019
a,\(-4x^2+4x-1\)
\(\Leftrightarrow\left(-2x-1\right)^2\)
b,\(\left(2x+1\right)^2-4\left(x-1\right)^2\)
\(\Rightarrow\left[2x+1-2\left(x-1\right)\right].\left[2x+1+2\left(x-1\right)\right]\)
\(\Rightarrow\left(2x+1-2x+2\right)\left(2x+1+2x-2\right)\)
\(\Rightarrow3\left(4x-1\right)\)
c,\(\left(2x-y\right)^2-4x^2+12x-9\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(4x^2-12x+9\right)\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(2x-3\right)^2\)
\(\Leftrightarrow\left(2x+y-2x+3\right)\left(2x+y+2x-3\right)\)
\(\Rightarrow\left(y+3\right)\left(4x+y-3\right)\)
d,\(\left(x+1\right)^2-4\left(x+1\right)y^2+4y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+2^2y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+4\left(y^2\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)-2y^2+\left(2y^2\right)^2\)
\(\Leftrightarrow\left(x+1-2y^2\right)^2\)
TQ
1
29 tháng 11 2018
\(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
14 tháng 8 2015
a) x^2 - 4 + ( x - 2 )^2
= ( x- 2 )(x + 2 ) + ( x- 2)^2
= ( x - 2 ) ( x + 2 + x - 2 )
= 2x (x-2)
b) x^3 - 2x^2 + x - xy^2
= x ( x^2 - 2x + 1 - y^2)
= x [ ( x - 1 )^2 - y^2 ]
= x(x - 1 - y)( x - 1 + y )
c) x^3 - 4x^2 - 12x + 27
= x^3 + 3x^2 - 7x^2 - 21x + 9x + 27
= x^2 ( x + 3 ) - 7x ( x+ 3 ) + 9(x + 3 )
Để hai lần nha
= ( x+ 3 )(x^2 - 7x + 9 )
30 tháng 9 2018
\(x^2-4+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=2x\left(x-2\right)\)
hk tốt
^^
HB
8 tháng 8 2016
a, x^2 -4+ (x-2)^2=(x-2)(x+2)+(x-2)^2=(x-2)(x+2+x-2)=(x-2)2x , b, x^3-2x^2+x-xy^2=x(x^2-2x+1-y^2)=x((x-1)^2-y^2)=x(x-1-y)(x-1+y) c,x^3-4x^2-4x^2-12x+27=(x^3+27)-(4x^2+12x)=(x+3)(x^2-3x+9)-4x(x+3)=(x+3)(x^2-7x+9) cách giải đó pn.......
18 tháng 11 2018
a) x2 - 4 + (x - 2)2
\(=\left(x^2-4\right)+\left(x-2\right)^2\)
\(=\left(x^2-2^2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left[\left(x+2\right)+\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
b) x3 - 2x2 + x - xy2
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left[\left(x-1-y\right)\left(x-1+y\right)\right]\)
\(=x\left(x-1-1\right)\left(x-1+y\right)\)
c) x3 - 4x2 - 12x + 27
\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x^3+3^3\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left[\left(x^2-3x+9\right)-4x\right]\)
\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
ND
0
Y
4
27 tháng 7 2016
\(x^3+2x^2+x-4xy^2\)
\(=x\left(x^2+2x+1\right)-4xy^2\)
\(=x\left(x+1\right)^2-4xy^2\)
\(=x\left(\left(x+1\right)^2-4y^2\right)\)
\(=x\left(\left(x+1-2y\right)\left(x+1+2y\right)\right)\)
27 tháng 7 2016
\(\text{x3+2x2+x−4xy2 =x(x2+2x+1)−4xy2 =x(x+1)2−4xy2 =x((x+1)2−4y2) =x((x+1−2y)(x+1+2y))}\)
a) x2 - 4 + (x - 2)2 = 0
=> (x - 2)(x + 2) + (x2 - 4x + 4) = 0
x2 + 2x - 2x - 4 + x2 - 4x + 4 = 0
2x2 - 4x = 0
2x(x - 2) = 0
=> 2x = 0 => x = 0
x - 2 = 0 x = 2
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