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TN
2
17 tháng 8 2021
Ta có: \(1+6x-6x^2-x^3\)
\(=-x^3-6x^2+6x+1\)
\(=\left(-x^3+1\right)-6x\left(x-1\right)\)
\(=-\left(x-1\right)\left(x^2+x+1\right)-6x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\left(-x^2-x-1-6x\right)\)
\(=-\left(x-1\right)\left(x^2+7x+1\right)\)
C
2
H
23 tháng 8 2023
1)
\(y^2-4y+4-x^2\\ =\left(y-2\right)^2-x^2\\ =\left(y-2-x\right)\left(y-2+x\right)\)
2)
\(8x^3-12x^2+6x-2\\ =2\left(4x^3-6x^2+3x-1\right)\\ =2\left(4x^3-4x^2-2x^2+2x+x-1\right)\\ =2\left(4x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\right)\\ =2\left(x-1\right)\left(4x^2-2x+1\right)\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
1) \(y^2-4y+4-x^2\)
\(=\left(y^2-4y+4\right)-x^2\)
\(=\left(y-2\right)^2-x^2\)
\(=\left(y-2-x\right)\left(y-2+x\right)\)
2) \(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)
\(=\left(2x-1\right)^3\)
\(=\left(2x-1\right)\left(2x-1\right)\left(2x-1\right)\)
M
1
19 tháng 11 2023
1: Đa thức này ko phân tích được nha bạn
2: \(x^2+8x+7\)
\(=x^2+x+7x+7\)
\(=x\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(x+7\right)\)
3: \(x^2-6x-16\)
\(=x^2-8x+2x-16\)
\(=x\left(x-8\right)+2\left(x-8\right)\)
\(=\left(x-8\right)\left(x+2\right)\)
4: \(4x^2-8x+3\)
\(=4x^2-2x-6x+3\)
\(=2x\left(2x-1\right)-3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x-3\right)\)
5: \(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
MH
3 tháng 10 2021
1) \(2xy^3-6x^2+10xy\)
\(=2x.y^3-2x.3x+2x.5y\)
\(=2x\left(y^3-3x+5y\right)\)
\(=2x[y\left(y^2-5\right)-3x]\)
NT
11 tháng 8 2021
1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)
2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)
3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
12 tháng 8 2021
1) Ta có: \(x^3+2x^2-6x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
2: Ta có: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
29 tháng 11 2023
bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)
\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)
Bài 2:
1: \(x^2y^2-8-1\)
\(=x^2y^2-9\)
\(=\left(xy-3\right)\left(xy+3\right)\)
2: \(x^3y-2x^2y+xy-xy^3\)
\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)
\(=xy\left(x^2-2x+1-y^2\right)\)
\(=xy\left[\left(x-1\right)^2-y^2\right]\)
\(=xy\left(x-1-y\right)\left(x-1+y\right)\)
3: \(x^3-2x^2y+xy^2\)
\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)
\(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
4: \(x^2+2x-y^2+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
5: \(x^2+2x-4y^2+1\)
\(=\left(x^2+2x+1\right)-4y^2\)
\(=\left(x+1\right)^2-4y^2\)
\(=\left(x+1-2y\right)\left(x+1+2y\right)\)
6: \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
21 tháng 8 2021
\(-9x^2+6x+y^2-1\)
\(=-\left(9x^2-6x+1-y^2\right)\)
\(=-\left(3x-1-y\right)\left(3x-1+y\right)\)
22 tháng 11 2017
a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)
b. \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)
c. \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)
T
11 tháng 12 2023
a) $x^3-3x^2y+4x-12y$
$=(x^3-3x^2y)+(4x-12y)$
$=x^2(x-3y)+4(x-3y)$
$=(x-3y)(x^2+4)$
b) $4x^2-y^2+4y-4$
$=4x^2-(y^2-4y+4)$
$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$
$=(2x)^2-(y-2)^2$
$=[2x-(y-2)][2x+(y-2)]$
$=(2x-y+2)(2x+y-2)$
c) $9x^2-6x-y^2+2y$
$=(9x^2-y^2)-(6x-2y)$
$=[(3x)^2-y^2]-2(3x-y)$
$=(3x-y)(3x+y)-2(3x-y)$
$=(3x-y)(3x+y-2)$
$\text{#}Toru$
1. \(=3x\left(2x+5\right)\)
2. \(=\left(3x-1\right)\left(3x+1\right)\)
3. \(=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
1, = 3x.(2x + 5)
2. = (3x)2 - 12 = (3x - 1).(3x +1 )
3, =(x2 + 6x + 9) - y2 = (x + 3)2 - y2 =(x + y -3 ). (x - y +3)