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Nguyên Hoàng
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Tên ?
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Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:17

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: \(3x^2-3y^2-2\left(x-y\right)^2\)

\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)

\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)

\(=\left(x-y\right)\left(x+5y\right)\)

Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:11

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: \(x^2\left(x+2y\right)-x-2y\)

\(=\left(x+2y\right)\left(x^2-1\right)\)

\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)

Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:12

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: \(x^4+2x^3+2x-1\)

\(=\left(x^2-1\right)\left(x^2+1\right)+2x\left(x^2+1\right)\)

\(=\left(x^2+1\right)\left(x^2+2x-1\right)\)

Lê Yến Vy
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Nguyễn Huy Tú
10 tháng 3 2022 lúc 18:49

1, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)

2, \(5x\left(3y+4x-6\right)\)

3, \(3x\left(2-y\right)\)

4, \(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)

5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)

6, \(2xy\left(x+2y-5x^2y\right)\)

7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

11, \(\left(x+y\right)\left(x-1\right)\)

Vô danh
10 tháng 3 2022 lúc 18:51

\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)

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Nguyễn Việt Lâm
15 tháng 12 2020 lúc 22:26

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.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

misen
1 tháng 7 2021 lúc 16:29

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-y= (x2-y2)(x2+y2)

 

 

misen
1 tháng 7 2021 lúc 16:36

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

 

Tên ?
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Trúc Giang
18 tháng 7 2021 lúc 16:36

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) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 22:59

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: \(x^3+2x^2+2x+1\)

\(=\left(x^3+1\right)+2x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+x+1\right)\)

Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:01

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: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)

\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

Triệu Việt Hà (Vịt)
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Akai Haruma
7 tháng 7 2021 lúc 20:29

a. $6x^2-11x=x(6x-11)$
b. $x^7+x^5+1=(x^7-x)+(x^5-x^2)+x+x^2+1$

$=x(x^6-1)+x^2(x^3-1)+(x^2+x+1)$
$=x(x^3-1)(x^3+1)+x^2(x^3-1)+(x^2+x+1)$
$=(x^3-1)(x^4+x+x^2)+(x^2+x+1)$

$=(x-1)(x^2+x+1)(x^4+x^2+x)+(x^2+x+1)$
$=(x^2+x+1)[(x-1)(x^4+x^2+x)+1]$

$=(x^2+x+1)(x^5-x^4+x^3-x+1)$

Akai Haruma
7 tháng 7 2021 lúc 20:34

c.

$x^8+x^4+1=(x^4)^2+2.x^4+1-x^4$

$=(x^4+1)^2-(x^2)^2$

$=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)(x^4+2x^2+1-x^2)$

$=(x^4-x^2+1)[(x^2+1)^2-x^2]$

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

d.

$x^3-5x+8-4=x^3-5x+4$

$=x^3-x^2+x^2-x-(4x-4)$

$=x^2(x-1)+x(x-1)-4(x-1)=(x-1)(x^2+x-4)$

e.

$x^5+x^4+1=(x^5-x^2)+(x^4-x)+x^2+x+1$

$=x^2(x^3-1)+x(x^3-1)+x^2+x+1$

$=(x^3-1)(x^2+x)+(x^2+x+1)$
$=(x-1)(x^2+x+1)(x^2+x)+(x^2+x+1)$

$=(x^2+x+1)[(x-1)(x^2+x)+1]$

$=(x^2+x+1)(x^3-x+1)$

 

 

Hưng Việt Nguyễn
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Lấp La Lấp Lánh
2 tháng 9 2021 lúc 10:44

Bài 2:

a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)

b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)

c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)

d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)

e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)

f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)

g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)

i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)

 

Nguyễn Lê Phước Thịnh
2 tháng 9 2021 lúc 13:52

a: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=\left(x+1\right)\left(3x-10\right)\)

b: \(x^2+6x+9-4y^2\)

\(=\left(x+3\right)^2-4y^2\)

\(=\left(x+3-2y\right)\left(x+3+2y\right)\)

c: \(x^2-2xy+y^2-5x+5y\)

\(=\left(x-y\right)^2-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-5\right)\)

nguyenduckhai /lop85
11 tháng 12 2021 lúc 9:35

a) 3x2−7x−10=(x+1)(3x−10)3x2−7x−10=(x+1)(3x−10)

b) x2+6x+9−4y2=(x+3)2−(2y)2=(x+3−2y)(x+3+2y)x2+6x+9−4y2=(x+3)2−(2y)2=(x+3−2y)(x+3+2y)

c) x2−2xy+y2−5x+5y=(x−y)2−5(x−y)=(x−y)(x−y−5)x2−2xy+y2−5x+5y=(x−y)2−5(x−y)=(x−y)(x−y−5)

d) 4x2−y2−6x+3y=(2x−y)(2x+y)−3(2x−y)=(2x−y)(2x+y−3)4x2−y2−6x+3y=(2x−y)(2x+y)−3(2x−y)=(2x−y)(2x+y−3)

e) 1−2a+2bc+a2−b2−c2=(a−1)2−(b−c)2=(a−1−b+c)(a−1+b−c)1−2a+2bc+a2−b2−c2=(a−1)2−(b−c)2=(a−1−b+c)(a−1+b−c)

f) x3−3x2−4x+12=(x+2)(x−3)(x−2)x3−3x2−4x+12=(x+2)(x−3)(x−2)

g) x4+64=(x2+8)2−16x2=(x2+8−4x)(x2+6+4x)x4+64=(x2+8)2−16x2=(x2+8−4x)(x2+6+4x)h) x4−5x2+4=(x+2)(x+1)(x−1)(x−2)x4−5x2+4=(x+2)(x+1)(x−1)(x−2)

i) (x+1)(x+3)(x+5)(x+7)+16=(x2+8x+7)(x2+8x+15)+16=(x2+8x+7)2+8(x2+8x+7)+16=(x2+8x+11)2(x+1)(x+3)(x+5)(x+7)+16=(x2+8x+7)(x2+8x+15)+16=(x2+8x+7)2+8(x2+8x+7)+16=(x2+8x+11)2

 

Mun SiNo
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Nguyễn Lê Phước Thịnh
11 tháng 10 2021 lúc 21:31

a: \(x^2-y^2-x-y\)

\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-1\right)\)

f: \(x^3-5x^2-5x+1\)

\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-6x+1\right)\)

Pham Trong Bach
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Cao Minh Tâm
2 tháng 9 2018 lúc 4:39