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a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
Bài 1:
a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)
b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]
= 2xy.(x-y-1).(x+y+1)
c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2
= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)
Bài 2:
a) (x+2).(x^2-2x+4) - (x^3+2x) = 0
x^3 + 8 - x^3 - 2x = 0
8 - 2x = 0
x = 4
b) x^2 - 2x - 8 = 0
x^2 +2x - 4x - 8 = 0
x.(x+2) - 4.(x+2) = 0
(x+2).(x-4) = 0
...
bn tự làm tiếp nha
1. \(x\left(x^2-5xy-14y^2\right)=x\left(x^2-7xy+2xy-14y^2\right)\)
\(=x\left(x-2\right)\left(x-7\right)\)
2. \(x^4+2x^2+1-9x^2=\left(x^2+1\right)^2-\left(3x\right)^2=\left(x^2+1-3x\right)\left(x^2+1+3x\right)\)
3. \(4x^4+4x^2+1-16x^2=\left(2x^2+1\right)^2-\left(4x\right)^2=\left(2x^2-4x+1\right)\left(2x^2+4x+1\right)\)
4. \(x^2+x+7x+7=\left(x+7\right)\left(x+1\right)\)
5. \(x\left(x^2-5x-14\right)=x\left(x^2-7x+2x-14\right)=x\left(x+2\right)\left(x-7\right)\)
Phân tích đa thức thành nhân tử :
1.x3-5x2y-14xy2
2.x4-7x2+1
3.4x4-12x2+1
4.x2+8x+7
5.x3-5x2-14x
Tạm thời phân tích như sau:
i) x4 - 2x3 + 2x - 1
= (x4 - 1) - (2x3 - 2x)
= (x2 + 1).(x2 -1) - 2x.(x2 - 1)
= (x2 - 1).(x2 - 2x + 1)
j) a6 - a4 + 2a3 + 2a2
= (a3 + a2).(a3 - a2) + 2.(a3 + a2)
= (a3 + a2).(a3 - a2 +2)
k) x4 - x3 + 2x2 + x + 1 (tạm thời giải thế này)
= x3.(x - 1) + (2x + 3 - \(\frac{4}{x-1}\)).(x -1)
= (x - 1).(x3 + 2x + 3 - \(\frac{4}{x-1}\))
Nếu đề là:
x4 + x3 + 2x2 + x + 1
= x4 + x2 + x3 + x + x2 + 1
= x2.(x2 + 1) + x.(x2 + 1) + x2 + 1
= (x2 + 1).(x2 + x + 1)
m) x2y + xy2 + x2z + y2z + 2xyz
= xy.(x + y) + z.(x2 + 2xy + y2)
= xy.(x + y) + z.(x + y).(x + y)
= (x + y).(xy + xz + yz)
n) x5 + x4 + x3 + x2 + x + 1
= x4.(x + 1) + x2.(x + 1) + x + 1
= (x + 1).(x4 + x2 + 1)
\(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-1\right)^2\)
\(=\left(x-1\right)^3\left(x+1\right)\)
Phân tích đa thức thành nhân tử
a) (1-2x)(1+2x)-x(x+2)(x-2)
\(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=\left(1-x^3\right)+\left(4x-4x^2\right)\)
\(=\left(1-x\right)\left(1+x+x^2\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+x+x^2+4x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8b^3a-8a^3b-12a^2b^2+6ab^3-b^4\)
\(=a^4+6a^3b+8b^3a-8a^3b-6ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(6a^3b-6ab^3\right)+\left(8b^3a-8a^3b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a^2-b^2\right)+8ab\left(b^2-a^2\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a-b\right)\left(a+b\right)-8ab\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3+6a^2b+6ab^2-8a^2b-8ab^2\right)\)
\(=\left(a-b\right)\left(a^3-a^2b-ab^2+b^3\right)\)
\(=\left(a-b\right)\left[a^2\left(a-b\right)-b^2\left(a-b\right)\right]\)
\(=\left(a-b\right)^3\left(a+b\right)\)
1, 2x2 - 8xy - 5x + 20y
= (2x2 - 5x) - (8xy - 20y)
= x(2x - 5) - 4y(2x - 5)
= (2x - 5) (x - 4y)
2, x3 - x2y - xy + y2
= (x3 - xy) - (x2y - y2)
= x(x2 - y) - y(x2 - y)
= (x2 - y) (x - y)
3, x2 - 2xy - 4z2 + y2
= (x2 - 2xy + y2) - 4z2
= (x - y)2 - (2z)2
= (x - y - 2z) (x - y + 2z)
4, a3 + a2b - a2c - abc
= (a3 - a2c) + (a2b - abc)
= a2(a - c) + ab(a - c)
= (a - c) (a2 + ab)
5, x3 + y3 + 3x2y + 3xy2 - x - y
= (x3 + 3x2y + 3xy2 + y3) - (x + y)
= (x + y) 3 - (x + y)
= (x + y) [(x + y)2 - 1]
= (x + y) (x + y - 1) (x + y + 1)