Phân tích đa thức thành nhân tử:
x4 - 2x3 - 2x2 - 2x -3
Phân tích các đa thức sau thành nhân tử: x 4 - 2 x 3 - 2 x 2 - 2 x - 3
x 4 - 2 x 3 - 2 x 2 - 2 x - 3 = ( x 4 − 1 ) − ( 2 x 3 + 2 x 2 ) − ( 2 x + 2 ) = ( x 2 + 1 ) ( x 2 − 1 ) − 2 x 2 ( x + 1 ) − 2 ( x + 1 ) = ( x 2 + 1 ) ( x − 1 ) ( x + 1 ) − 2 x 2 ( x + 1 ) − 2 ( x + 1 ) = ( x + 1 ) ( x 2 + 1 ) ( x − 1 ) − 2 x 2 – 2 = ( x + 1 ) ( x 2 + 1 ) ( x − 1 ) − 2 ( x 2 + 1 ) = ( x + 1 ) ( x 2 + 1 ) ( x – 1 − 2 ) = ( x + 1 ) ( x 2 + 1 ) ( x − 3 )
x^4 - 2x^3 - 2x^2 - 2x - 3
= x^4 - 1 - 2x^3 - 2x^2 - 2x -2
= ( x - 1 ) ( x + 1 ) ( x^2 + 1 ) - 2x^2 ( x + 1 ) - 2 ( x + 1 )
= ( x + 1 ) [ ( x - 1 ) ( x^2 + 1 ) - 2x^2 - 2 ]
= ( x + 1 ) [ ( x - 1 ) ( x^2 + 1 - 2 ( x^2 - 1 ) ]
= ( x + 1 ) [ ( x - 1 ) ( x^2 + 1 ) - 2 ( x - 1 ) ( x + 1 ) ]
= ( x + 1 ) ( x - 1 ) [ ( x^2 + 1 ) - 2 ( x +1 )
= ( x + 1 ) ( x - 1 ) ( x^2 +1 - 2x - 2 )
= ( x + 1 ) ( x - 1 ) ( x^2 - 2x - 1 )
Phân tích các đa thức sau thành nhân tử
a,x4+2x3+3x2+2x+1
b,x4-4x3+2x2+4x+1
c,x4+x3+2x2+2x+4
phân tích đa thức thành nhân tử
x4-2x3+2x-1
x⁴ - 2x³ + 2x - 1
= (x⁴ - 1) - (2x³ - 2x)
= (x² - 1)(x² + 1) - 2x(x² - 1)
= (x² - 1)(x² + 1 - 2x)
= (x - 1)(x + 1)(x² - 2x + 1)
= (x - 1)(x + 1)(x - 1)²
= (x - 1)³(x + 1)
Bài 1:phân tích đa thức thành nhân tử
a)x2-2x-4y2-4y e)x4+2x3+2x2+2x+1
b)x3+2x2+2x+1 f)x5+x4+x3+x2+x+1
c)x3-4x2+12x-27
d)a6-a4+2a3+2a2
Làm chi tiết giúp mình với ạ, cảm ơn
a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
b) \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)
d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)
a) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
b) Ta có: \(x^3+2x^2+2x+1\)
\(=\left(x^3+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
d) Ta có: \(a^6-a^4+2a^3+2a^2\)
\(=a^2\left(a^4-a^2+2a+2\right)\)
\(=a^2\left[a^2\left(a^2-1\right)+\left(2a+2\right)\right]\)
\(=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]\)
\(=a^2\cdot\left(a+1\right)\left(a^3-a+2\right)\)
c) Ta có: \(x^3-4x^2+12x-27\)
\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
Bài 1: phân tích đa thức thành nhân tử
a)x2-y2-2x-2y e)x4-2x3+2x-1
b)x2(x+2y)-x-2y f)x4+x3+2x2+x+1
c)x3-4x2-9x+36 g)x2y+xy2+x2z+y2z+2xyz
d)x4+2x3+2x-1 h)3x3-3y2-2(x-y)2
Làm chi tiết giúp mình với ạ , cảm ơn
e) Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\cdot\left(x-1\right)^3\)
h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
a) Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
c) Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
d) Ta có: \(x^4+2x^3+2x-1\)
\(=\left(x^2-1\right)\left(x^2+1\right)+2x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+2x-1\right)\)
phân tích đa thức thành nhân tử
x4 - 2x3 -2x2 -2x -3
x4 - 2x3-2x2 -2x -3
=(x4+x3)-(3x3+3x2)+(x2+x)-(3x+3)
=x3(x+1)-3x2(x+1)+x(x+1)-3(x+1)
= (x3-3x2+x-3)(x+1)
= ((x3-3x2)+(x-3))(x+1)
= (x2(x-3)+(x-3))(x+1)
=(x2+1)(x-3)(x+1)
Phân tích các đa thức sau thành nhân tử:
a) 3x - 3y + x 2 - y 2 ; b) x 2 -4 x 2 y 2 + y 2 + 2xy
c) x 6 - x 4 + 2 x 3 + 2 x 2 ; d) x 3 - 3x 2 +3x - 1 - y 3 .
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
x4+2x3+x2-y2
x3+x2-2x-8
phân tích đa thức thành nhân tử
a/ $=x^2(x^2+2x+1)-y^2\\=[x(x+1)]^2-y^2\\=[x(x+1)-y][x(x+1)+y]\\=(x^2+x-y)(x^2+x+y)$
b/ $=(x^3-8)+(x^2-2x)\\=(x-2)(x^2+2x+4)+x(x-2)\\=(x-2)(x^2+2x+5)$
\(x^4+2x^3+x^2-y^2=x^2\left(x+1\right)^2-y^2\\ =\left[x\left(x+1\right)-y\right]\left[x\left(x+1\right)+y\right]\\ =\left(x^2+x-y\right)\left(x^2+x+y\right)\\ x^3+x^2-2x-8=x^3-2x^2+3x^2-6x+4x-8\\ =\left(x-2\right)\left(x^2+3x-4\right)\)
phân tích thành nhân tử
a) x6 – x4 + 2x3 + 2x2
x⁶ - x⁴ + 2x³ + 2x²
= x²(x⁴ - x² + 2x + 2)
= x²[(x⁴ - x²) + (2x + 2)]
= x²[x²(x² - 1) +2(x + 1)]
= x²[x²(x - 1)(x + 1) + 2(x + 1)]
= x²(x + 1)[x²(x - 1) + 2]
= x²(x + 1)(x³ - x² + 2)
= x²(x + 1)(x³ + x² - 2x² - 2x + 2x + 2)
= x²(x + 1)[(x³ + x²) - (2x² + 2x) + (2x + 2)]
= x²(x + 1)[x²(x + 1) - 2x(x + 1) + 2(x + 1)]
= x²(x + 1)²(x² - 2x + 2)
phân tích đa thức: x4 + 2x3 + 4x2 + 3x + 2 thành nhân tử
Ta có:
\(\left(x^4+2x^3-x-2\right)+\left(4x^2+4x+4\right)\)
\(=\left[\left(x^4+2x^3\right)-\left(x+2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left[x^3\left(x+2\right)-\left(x-2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+1\right)+4\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x-1\right)\left(x+2\right)+4\right]\)
\(=\left(x^2+x+1\right)\left(x^2+x+2\right)\)