x4 -2x3 -4x2 + 4x-3 phân tichs đa thức thành nhân tử
phân tích đa thức: x4 + 2x3 + 4x2 + 3x + 2 thành nhân tử
Ta có:
\(\left(x^4+2x^3-x-2\right)+\left(4x^2+4x+4\right)\)
\(=\left[\left(x^4+2x^3\right)-\left(x+2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left[x^3\left(x+2\right)-\left(x-2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+1\right)+4\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x-1\right)\left(x+2\right)+4\right]\)
\(=\left(x^2+x+1\right)\left(x^2+x+2\right)\)
Bài 2 Phân tích đa thức sau thành nhân tử
a. x4 + 2x3 − 4x − 4
b. x2(1 − x2) − 4 − 4x2
c. x2 + y2 − x2y2 + xy − x − y
d* a3 + b3 + c3 − 3abc
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
d) Ta có: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
PHÂN TÍCH CÁC ĐA THỨC SAU THÀNH NHÂN TỬ BẰNG PHƯƠNG PHÁP NHÓM NHIỀU HẠNG TỬ :
a) x2 -2x -4y2-4y
b) x4 + 2x3 - 4x -4
c) x3 + 2x2y -x -2y
d) 3x2 -3y2 -2(x-y)2
e) x3 -4x2 -9x +36
f) x2 -y2 -2x -2y
a: Ta có: \(x^2-4y^2-2x-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c: Ta có: \(x^3+2x^2y-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
e: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
f: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
Phân tích đa thức thành nhân tử
x4+2x3-4x-4
Định lý Bezout
\(x^4+2x^3-4x-4\)
\(=\left(x^2-2\right)\left(x^2+2\right)-2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2-2x+2\right)\)
Phân tích các đa thức sau thành nhân tử:
d ) x 4 + 2 x 3 - 4 x – 4
d) x4 + 2x3 - 4x – 4 = (x4 – 4) + (2x3 – 4x) = (x2 – 2)(x2 + 2) + 2x(x2 – 2)
= (x2 – 2)(x2 + 2 + 2x) = (x - √2)( x + √2)( x2 + 2 + 2x)
Bài 1: Phân tích các đa thức sau thành nhân tử
a)x2-y2-2x+2y e)x4+4y4
b)x2(x-1)+16(1-x) f)x4-13x2+36
c)x2+4x-y2+4 g) (x2+x)2+4x2+4x-12
d)x3-3x2-3x+1 h)x6+2x5+x4-2x3-2x2+1
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
Phân tích đa thức thành nhân tử bằng phối hợp nhiều phương pháp
a) x4-4x2-4x-1
b) 10x4y2-10x3y2-10x2y2+10xy2
a) \(x^4-4x^2-4x-1=\left(x^4-1\right)-4x\left(x+1\right)=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-4x\left(x+1\right)=\left(x+1\right)\left[\left(x^2+1\right)\left(x-1\right)-4x\right]=\left(x+1\right)\left(x^3-x^2+x-1-4x\right)=\left(x+1\right)\left(x^3-x^2-3x-1\right)\)
b) \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2=10xy^2\left(x^3-x^2-x+1\right)=10xy^2\left(x-1\right)^2\left(x+1\right)\)
a: \(x^4-4x^2-4x-1\)
\(=\left(x^4-1\right)-4x\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-4x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x-x^2-1-4x\right)\)
\(=\left(x+1\right)\left(x^3-x^2-3x-1\right)\)
b: \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2\)
\(=10xy^2\left(x^3-x^2-x+1\right)\)
\(=10xy^2\cdot\left[\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\right]\)
\(=10xy^2\cdot\left(x+1\right)\left(x-1\right)^2\)
Bài 1: Phân tích các đa thức sau thành nhân tử
a. 1 - 4x2
b. 8 - 27x3
c. 27 + 27x + 9x 2 + x3
d. 2x3 + 4x2 + 2x
e. x2 - 5x - y2 + 5y
f. x2 - 6x + 9 - y2
g. 10x (x - y) - 6y(y - x)
h. x2 - 4x - 5
i. x4 - y4
Bài 2: Tìm x, biết
a. 5(x - 2) = x - 2
b. 3(x - 5) = 5 - x
c. (x +2)2 - (x+ 2) (x - 2) = 0
Bài 3: Tìm giá trị nhỏ nhất của biểu thức
a. A = x2 - 6x + 11
b. B = 4x2 - 20x + 101
c. C = -x2 - 4xy + 5y2 + 10x - 22y + 28
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
g. 10x(x-y)-6y(y-x)
=10x(x-y)+6y(x-y)
=(x-y)(10x+6y)
h.x2-4x-5
=(x-5)(x+1)
i.x4-y4 = (x2-y2)(x2+y2)
B2.
a.5(x-2)=x-2
⇔5(x-2)-(x-2)=0
⇔4(x-2)=0
⇔x=2
b.3(x-5)=5-x
⇔3(x-5)+(x-5)=0
⇔4(x-5)=0
⇔x=5
c.(x+2)2-(x+2)(x-2)=0
⇔(x+2)[(x+2)-(x-2)]=0
⇔4(x+2)=0
⇔x=-2
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 )