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27 tháng 10 2021

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

27 tháng 9 2023

a) x⁴ + 2x² + 1

= (x²)² + 2.x².1 + 1²

= (x² + 1)²

b) 4x² - 12xy + 9y²

= (2x)² - 2.2x.3y + (3y)²

= (2x - 3y)²

c) -x² - 2xy - y²

= -(x² + 2xy + y²)

= -(x + y)²

d) (x + y)² - 2(x + y) + 1

= (x + y)² - 2.(x + y).1 + 1²

= (x - y + 1)²

27 tháng 9 2023

e) x³ - 3x² + 3x - 1

= x³ - 3.x².1 + 3.x.1² - 1³

= (x - 1)³

g) x³ + 6x² + 12x + 8

= x³ + 3.x².2 + 3.x.2² + 2³

= (x + 2)³

h) x³ + 1 - x² - x

= (x³ + 1) - (x² + x)

= (x + 1)(x² - x + 1) - x(x + 1)

= (x + 1)(x² - x + 1 - x)

= (x + 1)(x² - 2x + 1)

= (x + 1)(x - 1)²

k) (x + y)³ - x³ - y³

= (x + y)³ - (x³ + y³)

= (x + y)³ - (x + y)(x² - xy + y²)

= (x + y)[(x + y)² - x² + xy - y²]

= (x + y)(x² + 2xy + y² - x² + xy - y²)

= (x + y).3xy

= 3xy(x + y)

27 tháng 10 2023

a, Sửa đề:

\(3x^2-\sqrt3 x+\dfrac14(dkxd:x\geq0)\\=(x\sqrt3)^2-2\cdot x\sqrt3\cdot\dfrac12+\Bigg(\dfrac12\Bigg)^2\\=\Bigg(x\sqrt3-\dfrac12\Bigg)^2\)

b, 

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

c,

\(x^4+x^3+2x^2+x+1\\=(x^4+x^3+x^2)+(x^2+x+1)\\=x^2(x^2+x+1)+(x^2+x+1)\\=(x^2+x+1)(x^2+1)\)

d,

\(x^3+2x^2+x-16xy^2\\=x(x^2+2x+1-16y^2)\\=x[(x+1)^2-(4y)^2]\\=x(x+1-4y)(x+1+4y)\\Toru\)

9 tháng 12 2023

a) x² - 9

= x² - 3²

= (x - 3)(x + 3)

b) 4x² - 1

= (2x)² - 1²

= (2x - 1)(2x + 1)

c) x⁴ - 16

= (x²)² - 4²

= (x² - 4)(x² + 4)

= (x² - 2²)(x² + 4)

= (x - 2)(x + 2)(x + 4)

d) x² - 4x + 4

= x² - 2.x.2 + 2²

= (x - 2)²

e) x³ - 8

= x³ - 2³

= (x - 2)(x² + 2x + 4)

f) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

25 tháng 12 2018

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 ].

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

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.

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

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

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

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.

$x^6+2x^5+x^4-2x^3-2x^2+1$

$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$

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

31 tháng 7 2021

a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)

b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)

c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

a) Ta có: \(x^3+4x-5\)

\(=x^3-x+5x-5\)

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

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

b) Ta có: \(x^3-3x^2+4\)

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

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

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

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

c) Ta có: \(x^3+2x^2+3x+2\)

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

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

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

d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)

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

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

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

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

30 tháng 9 2018

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

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

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

hk tốt

^^

23 tháng 11 2016

dài thế ai trả lời đc hả ?

23 tháng 11 2016

tu lam di luoi vua thoi

10 tháng 10 2021

a) \(=\left(x-2\right)^2\)

b) \(=\left(2x+1\right)^2\)

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

d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)

e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)

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

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

h) \(=\left(x+2\right)^3\)

i) \(=\left(1-x\right)^3\)

10 tháng 10 2021

a/ $=(x-2)^2$

b/ $=(2x+1)^2$

c/ $=(4x-3y)(4x+3y)$

d/ $=(1-x)(x+7)$

e/ $=(-x+1)(5x-1)$

f/ $=(x-y)(x^2+xy+y^2)$

g/ $=(3+x)(9-3x+x^2)$

h/ $=(x+2)^3$

i/ $=(1-x)^3$