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Nguyễn Thảo Nguyên
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Nguyễn Hoàng Minh
12 tháng 7 2023 lúc 21:56

\(a,=\left(5x^3+10x\right)+\left(x^4-4\right)\\ =5x\left(x^2+2\right)+\left(x^2+2\right)\left(x^2-2\right)\\ =\left(x^2+2\right)\left(x^2+5x-2\right)\\ b,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(c,=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\\ d,=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\\ e,=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^{10}-x^7+x^5-x^4+x^3-x+1\right)\)

Nguyễn Lê Phước Thịnh
12 tháng 7 2023 lúc 21:54

a: =x^4+2x^2+5x^3+10x-2x^2-4

=(x^2+2)(x^2+5x-2)

b; =(x+y)^3+z^3-3xy(x+y)-3xyz

=(x+y+z)*(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)

=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)

c: =x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1

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

Nguyên Hoàng
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Nguyễn Việt Lâm
5 tháng 8 2021 lúc 15:03

a.

\(x^3-y^3+2x^2-2y^2\)

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

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

b.

\(x^3+1-x^2-x\)

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

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

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

Nguyễn Đức Thắng
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Kiều Vũ Linh
25 tháng 5 2023 lúc 21:18

x⁸ + x⁴ + 1

= x⁸ + 2x⁴ + 1 - x⁴

= (x⁴ + 1)² - x⁴

= (x⁴ + 1)² - (x²)²

= (x⁴ + 1 + x²)(x⁴ + 1 - x²)

= (x⁴ + x² + 1)(x⁴ - x² + 1)

Nguyên Hoàng
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NLT MInh
5 tháng 8 2021 lúc 8:54
Khánh Linh Đỗ
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Nguyễn Lê Phước Thịnh
12 tháng 10 2023 lúc 16:46

a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)

c: \(2x-1-x^2\)

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

d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

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

g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)

\(=\left(5-x\right)\left(5+3x\right)\)

h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)

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

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

i: \(=x^2y^2-4xy+4-3\)

\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)

k: \(=y^2-\left(x-1\right)^2\)

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

l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)

m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)

Lê Hà Ny
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Norad II
28 tháng 10 2021 lúc 13:57

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

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

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

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

c) \(y^2+x^2+2xy-16=x^2+2xy+y^2-16\)

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

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.

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

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

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

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

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

 

 

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: \(x^4+2x^3+2x-1\)

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

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

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: \(x^3-4x^2+12x-27\)

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

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

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

hoàng minh vũ
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Lấp La Lấp Lánh
25 tháng 8 2021 lúc 15:40

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

b) \(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25=\left(x^2+7x+11\right)^2-5^2=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Nhan Thanh
25 tháng 8 2021 lúc 15:53

a. \(x^2\left(x^2+4\right)-x^2-4\)

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

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

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

b. \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

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

c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (*)

Đặt \(t=x^2+7x+10\), ta được

(*) \(=t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

hay \(\left(x^2+7x+6\right)\left(x^2+7x+18\right)\)

 

Nguyễn Lê Phước Thịnh
26 tháng 8 2021 lúc 1:12

a: Ta có: \(x^2\left(x^2+4\right)-x^2-4\)

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

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

b: Ta có: \(\left(x^2+x\right)^2+4x^2+4x-12\)

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

\(=\left(x^2+x\right)^2+6\left(x^2+x\right)-2\left(x^2+x\right)-12\)

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

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

c: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)